Chapter 4 · Dynamics: Force and Newton's Laws
Chapter 6 · Uniform Circular Motion and Gravitation
Chapter 7 · Work and Energy
Chapter 8 · Linear Momentum and Collisions
Chapter 11 · Fluid Statics
Chapter 12 · Fluid Dynamics
Chapter 13 · Temperature, Kinetic Theory, and the Gas Laws
Chapter 14 · Heat and Heat Transfer Methods
Chapter 15 · Thermodynamics
Chapter 16 · Oscillatory Motion and Waves
Chapter 17 · Physics of Hearing
Chapter 4: Dynamics: Force and Newton's Laws
4.1 Development of Force Concept
A force is a push or pull with both size and direction, so it's a vector and adds like any other vector.
- Dynamics — the study of the forces that cause motion.
- Force — a push or pull; has magnitude and direction.
- Forces add head-to-tail or with trig (components), just like velocity or displacement vectors.
- Free-body diagram — the object drawn as a single dot, with arrows for every external force on it. Internal forces are left off because they don't change the object's motion.
- Restoring force — the pull of a stretched spring back toward its relaxed length. A spring scale uses this to measure force.
Practice 4.1 in the quiz → · Top ↑
4.2 Newton's First Law: Inertia
An object keeps doing what it's doing (staying still, or moving in a straight line at constant speed) unless a net external force acts on it.
- First law — a body at rest stays at rest, and a body in motion keeps a constant velocity, unless acted on by a net external force.
- "Change in motion" means any change in velocity: speeding up, slowing down, or changing direction.
- Things slow down in real life because of friction, an external force. With zero friction (think air hockey), a sliding object would never slow.
- Inertia — an object's resistance to changes in its motion. The first law is also called the law of inertia.
- Mass — the measure of inertia, or the amount of matter. Measured in kg. It does not change with location (same on Earth, in orbit, on the Moon).
- Check: a kilogram of cotton and a kilogram of gold have the same mass (1 kg each). The gold is just smaller.
Practice 4.2 in the quiz → · Top ↑
4.3 Newton's Second Law: Concept of a System
Acceleration equals net external force divided by mass: push harder and it speeds up more, add mass and it speeds up less.
Key terms
- System — the object or group of objects you choose to study. Draw its boundary first.
- External force — comes from outside the system. Only these change the system's motion.
- Internal force — acts between parts of the system. Internal forces cancel in pairs, so ignore them.
- Net external force (F_net) — the vector sum of all external forces.
- Newton (N) — the SI unit of force: 1 N = 1 kg·m/s². In US units, 1 N ≈ 0.225 lb.
Weight vs. mass
|
Mass |
Weight |
| What it is |
Amount of matter (inertia) |
Force of gravity on the object |
| Unit |
kilogram (kg) |
newton (N) |
| Changes with location? |
No |
Yes |
| 1.0 kg object on Earth |
1.0 kg |
9.8 N |
| 1.0 kg object on the Moon (g = 1.625 m/s²) |
1.0 kg |
about 1.6 N |
- Free fall — when weight is the only force acting. Astronauts' "weightlessness" is really free fall, not zero gravity.
- A bathroom scale measures weight (a force) but is calibrated to display mass.
Worked examples
| Problem |
Given |
Setup |
Answer |
| Lawn mower (Ex. 4.1) |
F_net = 51 N, m = 24 kg |
a = F_net / m |
2.1 m/s² |
| Rocket sled (Ex. 4.2) |
4 rockets, m = 2100 kg, a = 49 m/s², f = 650 N |
4T − f = ma, so 4T = ma + f = 1.0×10⁵ N |
Each rocket: 2.6×10⁴ N |
Common mistakes
- Calling mass "weight" or giving weight in kg.
- Including internal forces in F_net.
- Using one force instead of the net force in F = ma.
Practice 4.3 in the quiz → · Top ↑
4.4 Newton's Third Law: Symmetry in Forces
Forces always come in pairs: if A pushes on B, B pushes back on A with the same size force in the opposite direction.
- Third law — whenever one body exerts a force on a second, the second exerts an equal and opposite force on the first. ("Action–reaction.")
- The pair never cancels, because the two forces act on different objects. Only forces on the same system can cancel.
- You move by pushing the opposite way: a swimmer pushes the wall backward and the wall pushes her forward.
Everyday examples
| You push on... |
...which pushes back on you |
Result |
| Floor, backward (walking) |
You, forward |
You walk forward |
| Road, backward (car tires) |
Car, forward |
Car accelerates |
| Exhaust gas, backward (rocket) |
Rocket, forward (thrust) |
Rocket accelerates, even in a vacuum |
| Air, downward (helicopter, bird wings) |
Aircraft, upward (lift) |
It flies |
| Ball, forward (bat swing) |
Bat, backward |
Bats can break, as with Mariano Rivera's cutter |
Choosing the system matters
The same situation gives different free-body diagrams depending on what you call the system. Pick the system so the force you want is external to it.
Worked examples
A professor (65.0 kg) pushes a cart (12.0 kg) of equipment (7.0 kg). She pushes back on the floor with 150 N; friction totals 24.0 N.
| Problem |
System |
Setup |
Answer |
| Acceleration (Ex. 4.3) |
Professor + cart + equipment (84 kg) |
Floor pushes her forward 150 N; F_net = 150 − 24.0 = 126 N; a = 126 / 84 |
1.5 m/s² |
| Force on the cart (Ex. 4.4) |
Cart + equipment (19.0 kg) |
F_net = ma = 19.0 × 1.5 = 29 N; F_prof = F_net + f = 29 + 24.0 |
53 N |
Only 53 N of her effort reaches the cart; the rest accelerates her own body.
Common mistakes
- Saying action and reaction cancel (they act on different objects).
- Thinking rockets push against the air or ground.
- Putting the force you exert on something onto your own free-body diagram.
Practice 4.4 in the quiz → · Top ↑
4.5 Normal, Tension, and Other Forces
A supporting surface pushes back perpendicular to itself (normal force), a rope or cable pulls along its length (tension), and on a slope you split weight into parallel and perpendicular parts.
Key terms
- Weight (w) — the force of gravity on an object: w = mg, always straight down.
- Normal force (N) — the support force a surface exerts, perpendicular ("normal") to the surface. Surfaces bend slightly and push back like a stiff spring.
- Tension (T) — the pull carried by a flexible connector (rope, cable, chain, tendon). It always acts along the connector, and it can only pull, never push.
- Real force — has a physical cause (gravity, a push, a rope).
- Fictitious force — only appears because the observer's frame is accelerating or rotating (e.g. the apparent sideways push on a merry-go-round).
- Inertial frame — a frame that isn't accelerating or rotating, where Newton's laws hold as written. Earth is close enough for almost all problems.
Normal force rules
- Flat surface, nothing else pushing or pulling vertically: N = mg.
- On an incline: N = mg cosθ, which is less than the weight.
- N can be greater than mg (pushing off when you jump, a car on a banked curve).
- Don't confuse N (normal force, a vector) with N (newton, the unit). A normal force is measured in newtons.
Inclines: resolving weight
Tilt the axes so one runs along the slope and one runs perpendicular to it. Then weight splits into two parts:
- w⊥ is balanced by N, so there's no motion into or off the slope.
- w∥ pulls the object down the slope; friction f points up the slope, opposing motion.
- Memory check: as θ → 0 (flat), sin → 0 and cos → 1, so all the weight presses into the surface. That's how you tell which is sin and which is cos.
Acceleration down an incline:
With no friction this becomes a = g sinθ, the same for any mass.
Tension rules
- A rope holding a hanging object at rest: T = w = mg.
- Tension is the same everywhere along an ideal (massless, frictionless) rope.
- Pulleys, tendons and brake cables change the direction of tension, not its size.
- A rope sagging at angle θ below horizontal, with a weight or push F⊥ at the middle:
- Small sag angle means huge tension. A rope can never be perfectly straight under load because that would need infinite tension. This is how you can pull a car out of mud by pushing sideways on a chain tied to a tree.
Worked examples
| Problem |
Given |
Setup |
Answer |
| Skier, no friction (Ex. 4.5a) |
m = 60.0 kg, θ = 25° |
a = g sin25° = 9.80 × 0.4226 |
4.14 m/s² |
| Skier, with friction (Ex. 4.5b) |
f = 45.0 N |
a = (mg sin25° − f) / m = (248.5 − 45.0) / 60.0 |
3.39 m/s² |
| Skier, normal force |
same |
N = mg cos25° = 60 × 9.8 × 0.906 |
532 N |
| Hanging mass |
m = 5.00 kg |
T = mg |
49.0 N |
| Tightrope walker (Ex. 4.6) |
m = 70.0 kg, sag 5.0° |
T = mg / (2 sin5.0°) = 686 / (2 × 0.0872) |
3,900 N (about 6× his weight) |
Problem-solving recipe
- Draw a free-body diagram: one dot, every force as an arrow.
- Pick axes that line up with the motion (along/perpendicular to a slope; horizontal/vertical for ropes).
- Break any force not on an axis into components with sin and cos.
- Write ΣF = ma separately for each axis. If nothing moves along an axis, ΣF = 0 there.
- Solve, then sanity-check units and size.
Common mistakes
- Assuming N = mg on a slope (it's mg cosθ).
- Swapping sin and cos on an incline.
- Forgetting tension acts on both sides of a sagging rope (the 2 in 2 sinθ).
- Mixing up the normal force N with the newton unit N.
- Leaving the calculator in radians instead of degrees.
Practice 4.5 in the quiz → · Top ↑
4.6 Problem-Solving Strategies
Every Newton's-laws problem follows the same four steps; the step students skip most is choosing the system and leaving out forces that don't act on it.
The four steps
- Sketch and identify forces. Draw the situation, then add an arrow for every force. Label each one, and make lengths and directions match the forces when you can.
- List knowns and unknowns, then define the system of interest. Draw a boundary around the object(s) you care about. Only external forces (from outside that boundary) go on the free-body diagram.
- Apply Newton's second law. Turn the free-body diagram into equations and solve.
- All forces parallel (1-D): add them like ordinary numbers with + and − signs.
- Forces at angles (2-D): split into two 1-D problems by projecting forces onto axes.
- Choose axes that make it easy: one axis along the direction of motion; on a ramp, along and perpendicular to the slope.
- Check the answer. Is it reasonable (e.g. friction should make an object slide slower, not faster)? Do the units match (a force should come out in newtons, not m/s)?
Free-body diagram rules
- Show only forces. No velocity or acceleration arrows.
- Show only external forces acting on the system. No internal forces.
- Use Newton's third law to sort them: a force the system exerts on something else belongs to that other object, not your diagram.
Tarzan example (Figure 4.20)
Tarzan hangs still from a vine. Three forces appear in the sketch: the vine's tension T pulling up on him, his weight w pulling down, and F_T, the force he exerts down on the vine. With Tarzan as the system, F_T acts on the vine, not on him, so it's left off the diagram. Since he's stationary, T and w balance: T = −w (same size, opposite direction).
Decide acceleration first, axis by axis
Before writing any equation, ask: is the system accelerating along this axis?
| Axis |
Accelerating? |
Equation |
| Direction of motion (say x) |
Yes |
F_net,x = ma |
| Perpendicular (say y) |
No |
F_net,y = 0 |
The zero-acceleration axis is usually where you find an unknown force like N or T.
Common mistakes
- Putting a force the object exerts on something else onto its own free-body diagram (like F_T on Tarzan).
- Drawing velocity or acceleration as if they were forces.
- Setting F_net = ma on an axis where nothing accelerates (it should be 0).
- Skipping the units check.
Practice 4.6 in the quiz → · Top ↑
4.7 Further Applications of Newton's Laws
This section stacks the earlier tools into harder problems: combine perpendicular forces, solve two unknown tensions with two equations, predict an elevator scale reading, and link F = ma to kinematics.
Combining perpendicular forces
When two forces are at right angles, find the total with the Pythagorean theorem and its direction with inverse tangent:
Drag (friction from a fluid like water or air) points opposite the motion. Once you know the direction of motion, the problem becomes 1-D: F_net = F_applied − F_drag = ma.
Unequal tensions
If a hanging object is held by two wires at different angles, the tensions differ. Use two equations (zero acceleration on both axes):
- Horizontal: T₁ cosθ₁ = T₂ cosθ₂ (horizontal pulls cancel)
- Vertical: T₁ sinθ₁ + T₂ sinθ₂ = mg (vertical pulls hold up the weight)
Solve the first for one tension, substitute into the second. The steeper (more vertical) wire carries more tension.
Elevator and bathroom scale
A scale reads the normal force it pushes up with. With the person as the system, F_s − mg = ma, so:
| Elevator motion |
a (up = +) |
Scale reads |
| Speeding up going up, or slowing down going down |
positive |
More than weight |
| At rest or constant speed (up or down) |
0 |
Exactly weight |
| Speeding up going down, or slowing down going up |
negative |
Less than weight |
| Free fall |
−g |
Zero ("weightless") |
Key idea: the scale responds to acceleration, not velocity. Constant speed feels the same as standing still.
Integrated problems: kinematics + forces
- List knowns and unknowns to spot which principles apply.
- Use kinematics (a = Δv / Δt, etc.) to find acceleration, or forces to find acceleration and then motion.
- Connect them through F_net = ma.
Worked examples
| Problem |
Given |
Setup |
Answer |
| Barge drag (Ex. 4.7) |
Tugs 2.7×10⁵ N (x) and 3.6×10⁵ N (y); m = 5.0×10⁶ kg; a = 0.075 m/s² |
F_app = √(2.7² + 3.6²)×10⁵ = 4.5×10⁵ N at 53°; F_D = F_app − ma |
7.5×10⁴ N, opposite the motion |
| Traffic light (Ex. 4.8) |
m = 15.0 kg; wires at 30° and 45° |
T₂ = 1.225 T₁; 1.366 T₁ = 147 N |
T₁ = 108 N, T₂ = 132 N |
| Scale, accelerating up (Ex. 4.9a) |
m = 75.0 kg; a = 1.20 m/s² |
F_s = 75.0(9.80 + 1.20) |
825 N (about 185 lb) |
| Scale, constant speed (Ex. 4.9b) |
v = 1 m/s, so a = 0 |
F_s = mg |
735 N (his weight) |
| Soccer sprint (Ex. 4.10) |
m = 70.0 kg; 0 → 8.00 m/s in 2.50 s |
a = 8.00 / 2.50 = 3.20 m/s²; F = ma |
224 N (about 50 lb) |
Common mistakes
- Assuming two wire tensions are equal when the angles differ.
- Thinking a scale reads more just because the elevator is moving up (only acceleration changes it).
- Subtracting drag in the wrong direction (it always opposes motion).
- Forgetting that the ground pushes the runner forward (Newton's third law) when he pushes it backward.
Practice 4.7 in the quiz → · Top ↑
Chapter 5: Friction
5.1 Friction
Friction opposes sliding between two surfaces and equals a coefficient times the normal force; it's stronger before things start moving (static) than once they slide (kinetic).
Two kinds of friction
- Static friction (f_s) — acts when surfaces aren't sliding. It matches your push, up to a maximum of μ_s N. Push harder than that and the object breaks loose.
- Kinetic friction (f_k) — acts while sliding. It's a fixed value, μ_k N, and is usually smaller than max static friction. That's why it's easier to keep a crate moving than to start it.
- Coefficient of friction (μ) — no units, usually between 0 and 1. Depends on the two materials, not on contact area.
- Direction: parallel to the surface, opposite the motion (or the attempted motion).
- More normal force (a heavier object, or pressing down) means more friction.
Selected coefficients (Table 5.1)
| Surfaces |
μ_s |
μ_k |
| Rubber on dry concrete |
1.0 |
0.7 |
| Shoes on wood |
0.9 |
0.7 |
| Rubber on wet concrete |
0.7 |
0.5 |
| Steel on steel (dry) |
0.6 |
0.3 |
| Wood on wood |
0.5 |
0.3 |
| Waxed wood on wet snow |
0.14 |
0.1 |
| Shoes on ice |
0.1 |
0.05 |
| Steel on steel (oiled) |
0.05 |
0.03 |
| Bone joint (synovial fluid) |
0.016 |
0.015 |
Friction on a slope
On an incline, N = mg cosθ, so f_k = μ_k mg cosθ. If an object slides down at constant velocity, friction exactly balances the downhill pull:
That gives a quick way to measure μ_k: tilt a surface until something slides at steady speed and take tan of the angle.
Worked examples
| Problem |
Given |
Setup |
Answer |
| Crate, to start moving |
m = 100 kg, μ_s = 0.45 |
N = mg = 980 N; f_s(max) = 0.45 × 980 |
Push over 440 N |
| Crate, to keep moving |
μ_k = 0.30 |
f_k = 0.30 × 980 |
290 N at constant speed |
| Skier on 25° slope (Ex. 5.1) |
m = 62 kg, f_k = 45.0 N |
μ_k = f_k / (mg cos25°) = 45.0 / (62 × 9.80 × 0.906) |
μ_k = 0.082 |
Why friction happens
Surfaces are rough up close, so only the high spots touch. Molecules at those spots also stick to each other (adhesion). Pressing harder increases the true contact area, which is why friction depends on normal force, not on the overall surface area.
Common mistakes
- Using μ_s N as the friction force when the object isn't on the verge of slipping (static friction only equals what's needed).
- Using N = mg on a slope instead of mg cosθ.
- Pointing friction in the direction of motion.
- Giving μ units (it has none).
Practice 5.1 in the quiz → · Top ↑
Chapter 6: Uniform Circular Motion and Gravitation
6.1 Rotation Angle and Angular Velocity
Angles in rotation are measured in radians, and angular velocity (ω) tells how fast the angle changes; a point's speed is v = rω.
- Rotation angle (Δθ) — how far something has turned. It's arc length divided by radius.
- Arc length (Δs) — distance traveled along the circle.
- Radius of curvature (r) — radius of the circular path.
- Radian (rad) — the unit for rotation angles. One revolution = 2π rad. Radians have no real units (meters ÷ meters), so rad/s works out as 1/s.
- Angular velocity (ω) — rate of rotation, in rad/s. Its direction is just clockwise or counterclockwise.
- Tangential (linear) speed (v) — speed of a point moving around the circle, pointing along the tangent.
Conversions
| From |
To |
Multiply by |
| Revolutions |
Radians |
2π |
| Degrees |
Radians |
π / 180 |
| Radians |
Degrees |
180 / π (1 rad ≈ 57.3°) |
| rev/min (rpm) |
rad/s |
2π / 60 |
Common angles: 30° = π/6, 60° = π/3, 90° = π/2, 120° = 2π/3, 135° = 3π/4, 180° = π, 360° = 2π.
Key ideas
- Every point on a spinning object turns through the same angle in the same time, so it has the same ω.
- Points farther from the center travel farther, so they move faster: v = rω. The tip of a bat moves faster than the handle.
- For the same road speed, a bigger wheel spins slower (ω = v/r).
- After one full revolution (2π rad), every point is back where it started.
Worked examples
| Problem |
Setup |
Answer |
| Car tire, r = 0.300 m, car at 15.0 m/s (Ex. 6.1) |
ω = v/r = 15.0 / 0.300 |
50.0 rad/s |
| Earth-mover tire, r = 1.20 m, same speed |
ω = 15.0 / 1.20 |
12.5 rad/s |
Common mistakes: using degrees in v = rω or Δs = rΔθ (must be radians); forgetting the 2π when converting revolutions; thinking the outer edge has a bigger ω (it has a bigger v, same ω); calculator in the wrong mode.
rpm conversion isn't in 6.1 itself, but it shows up in problems, so it's included above.
Practice 6.1 in the quiz → · Top ↑
6.2 Centripetal Acceleration
An object moving in a circle at constant speed is still accelerating, because its direction keeps changing; that acceleration points to the center.
- Centripetal acceleration (a_c) — "center-seeking." Always points toward the center of the circle, perpendicular to the velocity.
- It grows with the square of speed: twice as fast around the same curve needs 4× the acceleration.
- A tighter curve (smaller r) means more acceleration at the same speed.
- Use v²/r when you know speed; use rω² when you know angular velocity.
- Centrifuges spin samples so fast that a_c is thousands of g's, which separates blood cells, DNA and proteins quickly. Expressing a_c "in g's" means a_c / 9.80.
| Problem |
Setup |
Answer |
| Car on a 500 m curve at 25.0 m/s (Ex. 6.2) |
a_c = 25.0² / 500 |
1.25 m/s² (0.128g) |
| Ultracentrifuge, r = 0.0750 m, 7.50 × 10⁴ rpm (Ex. 6.3) |
ω = 7854 rad/s; a_c = 0.0750 × 7854² |
4.63 × 10⁶ m/s² (472,000g) |
Common mistakes: thinking constant speed means no acceleration; pointing a_c outward; forgetting to square v or ω; not converting rpm to rad/s.
Practice 6.2 in the quiz → · Top ↑
6.3 Centripetal Force
Whatever net force points to the center is the centripetal force; it isn't a new kind of force.
- Centripetal force — the net inward force causing circular motion. It's always supplied by a real force:
| Situation |
What supplies F_c |
| Car on a flat curve |
Static friction from the road |
| Car on a banked curve (ideal) |
Horizontal part of the normal force |
| Tetherball, ball on a string |
Tension |
| Moon or satellite in orbit |
Gravity |
| Skater turning |
Friction on the skates |
- More force at the same speed means a tighter circle: r = mv²/F_c.
- If the force disappears (string breaks, road too icy), the object flies off in a straight line tangent to the circle, not straight outward.
Flat curve: friction
This is the minimum coefficient needed. Mass cancels, so a loaded truck and a car need the same μ at the same speed.
Banked curve with no friction
Faster speeds and tighter curves need steeper banking. Mass cancels again. Friction lets cars go faster or slower than the ideal speed.
| Problem |
Setup |
Answer |
| 900 kg car, r = 500 m, 25.0 m/s: force (Ex. 6.4a) |
F_c = 900 × 25.0² / 500 |
1125 N |
| Minimum μ_s for that curve (Ex. 6.4b) |
μ_s = 25.0² / (500 × 9.80) |
0.13 |
| Ideal speed on a 65.0° bank, r = 100 m (Ex. 6.5) |
v = √(100 × 9.80 × tan65°) |
45.8 m/s (165 km/h) |
Common mistakes: drawing "centripetal force" as an extra force on a free-body diagram; thinking an object flies straight outward when released.
Practice 6.3 in the quiz → · Top ↑
6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
Forces you seem to feel in a turning or accelerating vehicle aren't real; they come from your own inertia.
- Inertial frame — not accelerating or rotating; every force has a real source. Earth is almost inertial.
- Non-inertial frame — accelerating or rotating (a turning car, a merry-go-round, a jet taking off).
- Fictitious force — an apparent force caused by the frame's motion, with no physical source.
- Centrifugal force — the fictitious outward "push" you feel on a merry-go-round. Really, your body is trying to go straight and the ride pushes you inward. Centripetal (real, inward) is not the same as centrifugal (fictitious, outward).
- Coriolis force — a fictitious force that makes moving objects seem to curve in a rotating frame. On Earth it deflects motion to the right in the Northern Hemisphere and left in the Southern.
- Hurricanes spin counterclockwise in the Northern Hemisphere (clockwise in the Southern). High-pressure systems spin the other way.
- The Coriolis effect matters for big things like weather and long-range artillery, not a thrown baseball.
Common mistakes: saying a "centrifugal force" throws you outward in a turning car; mixing up the hemisphere directions.
Practice 6.4 in the quiz → · Top ↑
6.5 Newton's Universal Law of Gravitation
Every mass attracts every other mass with a force that weakens with the square of the distance between their centers.
- r is measured center to center. On Earth's surface, r = Earth's radius.
- Inverse square: double the distance, ¼ the force; triple it, 1/9.
- g doesn't depend on the falling object's mass, so everything falls at the same rate.
- Earth pulls the Moon exactly as hard as the Moon pulls Earth (Newton's third law).
- Weightlessness in orbit is free fall, not zero gravity. Gravity at the space station is still about 90% of surface gravity.
- Tides: the Moon pulls the near side of Earth more than the center and the center more than the far side, making two bulges, so two high tides a day. Spring tides (biggest) when Sun, Earth and Moon line up; neap tides (smallest) when they're at 90°.
- Cavendish measured G in 1798, which let scientists calculate Earth's mass: M = gr²/G.
Useful values: Earth M = 5.98 × 10²⁴ kg, R = 6.38 × 10⁶ m; Earth–Moon distance 3.84 × 10⁸ m; Moon's period 27.3 days.
| Problem |
Setup |
Answer |
| Earth's gravity at the Moon's distance (Ex. 6.6a) |
g = GM / r² with r = 3.84 × 10⁸ m |
2.70 × 10⁻³ m/s² |
| Moon's centripetal acceleration (Ex. 6.6b) |
ω = 2π / 27.3 days = 2.66 × 10⁻⁶ rad/s; a_c = rω² |
2.72 × 10⁻³ m/s² (they match, so gravity holds the Moon in orbit) |
| Two 1.000 kg masses 1.000 m apart |
F = G × 1 × 1 / 1² |
6.674 × 10⁻¹¹ N (tiny) |
Common mistakes: measuring r from the surface instead of the center; forgetting to square r; saying there's no gravity in space.
Practice 6.5 in the quiz → · Top ↑
6.6 Satellites and Kepler's Laws
Kepler described how planets move; Newton's gravity explains why.
- First law: every orbit is an ellipse with the central body at one focus.
- Second law: the line from the Sun to a planet sweeps out equal areas in equal times, so the planet moves fastest when closest.
- Third law: T² is proportional to r³ for everything orbiting the same body.
- For a circular orbit, gravity is the centripetal force: GmM/r² = mv²/r. The satellite's mass cancels, so every satellite at a given radius has the same speed and period.
- Only compare orbits around the same central body with the third law.
- r is from the center of the planet: altitude + planet radius.
- Measuring a moon's r and T gives the planet's mass.
| Problem |
Setup |
Answer |
| Satellite 1500 km up, compared with the Moon (Ex. 6.7) |
r₂ = 1500 + 6380 = 7880 km; T₂ = 27.3 d × (7880 / 3.84 × 10⁵)^1.5 |
1.93 h |
Planets around the Sun all share r³/T² ≈ 3.35 × 10²⁴ km³/y².
Common mistakes: using altitude as r; comparing a moon of Jupiter with a planet around the Sun in Kepler's third law; thinking heavier satellites orbit faster.
Practice 6.6 in the quiz → · Top ↑
Chapter 7: Work and Energy
7.1 Work: The Scientific Definition
Work is done only when a force moves something, and only the part of the force along the motion counts.
- Work (W) — energy transferred by a force acting through a displacement. Measured in joules: 1 J = 1 N·m = 1 kg·m²/s². One joule lifts a 100 g apple about 1 m.
- θ is the angle between the force and the displacement.
| Situation |
Angle |
Work |
| Force along the motion (pushing a cart forward) |
0° |
Positive, W = Fd |
| Force at an angle (pushing a mower handle 35° below horizontal) |
θ |
W = Fd cosθ |
| Force perpendicular to motion (carrying a briefcase across a level floor) |
90° |
Zero |
| No displacement (holding a heavy box still) |
— |
Zero |
| Force opposite the motion (friction, lowering a load) |
180° |
Negative, W = −Fd |
- Positive work puts energy into the object; negative work takes energy out.
- You get tired holding a weight because your muscles do work on each other, not on the weight.
- Food calories: 1 kcal = 4186 J.
Worked example (Ex. 7.1): pushing a mower with 75.0 N at 35° below horizontal for 25.0 m: W = 75.0 × 25.0 × cos35° = 1.54 × 10³ J (0.367 kcal).
Common mistakes: using sin instead of cos; calling it work when the force is perpendicular to the motion; forgetting the negative sign for friction.
Practice 7.1 in the quiz → · Top ↑
7.2 Kinetic Energy and the Work-Energy Theorem
The net work done on an object equals its change in kinetic energy.
- Kinetic energy (KE) — energy of motion, in joules.
- KE grows with the square of speed: double the speed and KE is 4×. A car at 100 km/h has 4× the KE it has at 50 km/h.
- Net work can be found two ways: W_net = F_net × d, or add up the work done by each force. They give the same answer.
- Forces perpendicular to the motion (weight and normal force on a level surface) do zero work.
- Work is also the area under an F vs. d graph.
Worked examples (30.0 kg package on rollers)
| Problem |
Setup |
Answer |
| KE at 0.500 m/s (Ex. 7.2) |
½ × 30.0 × 0.500² |
3.75 J |
| Net work: 120 N push, 5.00 N friction, 0.800 m (Ex. 7.3) |
(120 − 5.00) × 0.800, or 96.0 J − 4.00 J |
92.0 J |
| Final speed (Ex. 7.4) |
½mv² = 92.0 + 3.75 = 95.75 J; v = √(2 × 95.75 / 30.0) |
2.53 m/s |
| Coasting distance after the push (Ex. 7.5) |
Friction removes 95.75 J: d = 95.75 / 5.00 |
19.2 m |
Common mistakes: forgetting to square v; using the applied force instead of the net force; leaving out the initial KE.
Practice 7.2 in the quiz → · Top ↑
7.3 Gravitational Potential Energy
Lifting something stores energy that comes back as kinetic energy when it falls.
- Gravitational potential energy (PE_g) — energy stored by height, in the object–Earth system.
- Only the change in height matters. You pick where h = 0 (usually the ground or the lowest point).
- Path doesn't matter: a roller coaster has the same speed at the bottom whether it drops straight down or follows a winding track (no friction).
- Mass cancels in the speed formula, so heavy and light objects reach the same speed falling the same height.
- Stopping over a short distance means a huge force: F = mgh / d.
Worked examples
| Problem |
Setup |
Answer |
| Raise a 0.500 kg clock weight 1.00 m |
0.500 × 9.80 × 1.00 |
4.90 J |
| Roller coaster from rest, 20.0 m drop (Ex. 7.7a) |
√(2 × 9.80 × 20.0) |
19.8 m/s |
| Same coaster starting at 5.00 m/s (Ex. 7.7b) |
√(392 + 25.0) |
20.4 m/s |
| 60.0 kg person lands stiffly from 3.00 m, knees bend 0.500 cm (Ex. 7.6) |
F = mgh / d = 1764 / 0.00500 |
3.53 × 10⁵ N (500× his weight) |
Note in 7.7b: starting at 5 m/s only raises the final speed from 19.8 to 20.4 m/s, not by 5 m/s, because energies add, not speeds.
Common mistakes: adding speeds instead of energies; forgetting h is the vertical drop, not the track length; keeping m when it cancels.
Practice 7.3 in the quiz → · Top ↑
7.4 Conservative Forces and Potential Energy
If only conservative forces act, kinetic plus potential energy stays constant.
- Conservative force — its work depends only on the start and end points, not the path. Gravity and springs are conservative; friction is not.
- Potential energy (PE) — stored energy due to position, shape, or configuration; fully recoverable.
- Spring potential energy — ½kx², where k is the spring constant (N/m) and x is the stretch or compression from the relaxed length. Hooke's law: F = kx.
- Double the compression and PE_s is 4×.
- Mechanical energy = KE + PE. It's conserved when friction is negligible.
Worked example (Ex. 7.8): a 0.100 kg toy car, spring with k = 250.0 N/m compressed 4.00 cm, track rises 0.180 m.
| Point |
Energy equation |
Speed |
| Leaving the spring |
½kx² = ½mv² → v = x√(k/m) = 0.0400 × √2500 |
2.00 m/s |
| Top of the 0.180 m rise |
½kx² = ½mv² + mgh → v = √(kx²/m − 2gh) |
0.687 m/s |
Common mistakes: using cm instead of m for x; forgetting to square x; using KE_i + PE_i = KE_f + PE_f when friction is significant.
Practice 7.4 in the quiz → · Top ↑
7.6 Conservation of Energy
Total energy never disappears; it only changes form or moves between systems.
- Law of conservation of energy — total energy is constant in any process.
- W_nc — work by nonconservative forces like friction (usually negative).
- OE (other energy) — chemical, thermal, electrical, nuclear, radiant (light), and so on.
- Thermal energy — the random motion of atoms and molecules; friction usually turns mechanical energy into thermal energy.
- Efficiency — the fraction of input energy that comes out as useful work. Always less than 100%; the rest mostly becomes thermal energy.
Energy chains
| Process |
Conversion |
| Eating and moving |
Chemical → kinetic + potential + thermal |
| Photosynthesis |
Light → chemical |
| Coal power plant |
Chemical → thermal → mechanical (turbine) → electrical |
| Solar-powered motor |
Radiant → electrical → mechanical |
Typical efficiencies (Table 7.2)
| Device or activity |
Efficiency |
| Electric motor |
98% |
| Gas heater |
90% |
| Coal power plant |
42% |
| Diesel engine |
35% |
| Gasoline engine |
30% |
| Cycling and climbing |
20% |
| Solar cell |
10% |
| Weightlifting |
9% |
| Shoveling |
3% |
| Surface swimming |
2% |
Energy problem-solving steps
- Define the system, knowns, and unknown. Sketch it.
- List the forces. Do you know a potential energy for each?
- All conservative → KE_i + PE_i = KE_f + PE_f.
- Friction or other energy involved → use the full equation with W_nc and OE.
- Zero out terms: set h = 0 at the start or end; cross out KE when something is at rest.
- Check: friction work should be negative, PE lower at the bottom, speeds reasonable.
Common mistakes: saying energy is "lost" (it's converted, usually to thermal); efficiency over 100%; calculating work by gravity separately and also counting PE (that double-counts it).
Section 7.5 (Nonconservative Forces) wasn't included in this batch. 7.6 uses W_nc from it, so it's worth adding if it's assigned.
Practice 7.6 in the quiz → · Top ↑
Chapter 8: Linear Momentum and Collisions
8.1 Linear Momentum and Force
Momentum is mass times velocity, and a net force is what changes it.
- Linear momentum (p) — a vector in the direction of the velocity. Units: kg·m/s.
- More mass or more speed means more momentum. A slow lineman can have far more momentum than a fast football.
- Newton's second law, momentum form: F_net = Δp/Δt. Newton wrote it this way. It reduces to F = ma when mass is constant, but it also works when mass changes (rockets).
- Signs matter: pick a positive direction. A ball that reverses direction has a change in momentum of m(v_f − v_i), which is bigger than mv.
| Problem |
Setup |
Answer |
| 110 kg player at 8.00 m/s (Ex. 8.1a) |
p = 110 × 8.00 |
880 kg·m/s |
| 0.410 kg football at 25.0 m/s (Ex. 8.1b) |
p = 0.410 × 25.0 |
10.3 kg·m/s (player has about 85× more) |
| Tennis serve: 0.057 kg, 0 → 58 m/s in 5.0 ms (Ex. 8.2) |
Δp = 0.057 × 58 = 3.3 kg·m/s; F = 3.3 / 0.0050 |
660 N |
Common mistakes: forgetting momentum's direction (sign); using m × final speed instead of m × change in velocity; ms not converted to s.
Practice 8.1 in the quiz → · Top ↑
8.2 Impulse
Impulse is force times time, and it equals the change in momentum.
- Impulse — units N·s, the same as kg·m/s.
- The same Δp can come from a big force for a short time or a small force for a long time.
- Safety: airbags, padded dashboards, crumple zones, bending your knees, and "giving" with your glove all increase Δt, which lowers the force for the same Δp.
- Force-time graphs: the impulse is the area under the curve. The average effective force is the constant force with the same area.
- Bouncing gives a bigger impulse than stopping. A ball that rebounds at the same speed has Δp = 2mv; one that stops has Δp = mv.
- At an angle, only the momentum component perpendicular to the wall changes (Ex. 8.3: a ball hitting at 30° gets impulse 2mu cos30°, so the head-on ball's is 1.155× larger).
Common mistakes: saying a cushion reduces impulse (it reduces force; the impulse is the same); forgetting the factor of 2 for a bounce.
Practice 8.2 in the quiz → · Top ↑
8.3 Conservation of Momentum
If no net external force acts on a system, its total momentum never changes.
- Isolated system — net external force is zero. Internal forces (the two objects pushing each other) cancel by Newton's third law.
- Primes (′) mean "after."
- One object's momentum can change, but the system total can't. Choose the system to include both objects.
- Recoil: things that start at rest and push apart end with equal and opposite momenta (rifle and bullet, two skaters, squid and water jet).
- Momentum can be conserved in one direction and not another: a projectile's horizontal momentum is conserved, but gravity changes its vertical momentum.
Common mistakes: dropping the sign of a velocity going the other way; including an object in the system that has an outside force on it (like a falling ball with gravity).
Practice 8.3 in the quiz → · Top ↑
8.4 Elastic Collisions in One Dimension
In an elastic collision both momentum and kinetic energy are conserved.
- Elastic collision — internal kinetic energy (the sum of all the KEs) is the same before and after.
- Perfectly elastic collisions happen mainly between atoms and subatomic particles. Air-track carts with spring bumpers and steel blocks on ice come close.
- Shortcut when m₂ starts at rest (follows from the two equations above):
| Case (target at rest) |
What happens |
| Equal masses |
The first object stops; the second leaves at the first one's speed (Newton's cradle) |
| Light hits heavy |
Light one bounces back; heavy one moves forward slowly |
| Heavy hits light |
Heavy one keeps going slightly slower; light one shoots forward at up to 2v |
Worked example (Ex. 8.4): 0.500 kg at 4.00 m/s hits 3.50 kg at rest. v₁′ = (0.500 − 3.50)/4.00 × 4.00 = −3.00 m/s and v₂′ = 2(0.500)/4.00 × 4.00 = 1.00 m/s. Check: KE = 4.00 J and p = 2.00 kg·m/s both before and after.
Practice 8.4 in the quiz → · Top ↑
8.5 Inelastic Collisions in One Dimension
In an inelastic collision momentum is still conserved, but kinetic energy is not.
- Inelastic collision — internal KE changes: some becomes thermal energy, sound, or bent metal. (Stored energy, like a spring or an explosion, can also add KE.)
- Perfectly inelastic — the objects stick together. This loses the most KE possible while still conserving momentum.
- Two equal masses at equal speeds hitting head-on and sticking stop dead: final momentum and final KE are both zero.
- Coefficient of restitution: c = √(h/H) for a ball dropped from H that bounces to h. A new tennis ball is about 0.85.
- In sports, a heavier bat or racquet can transfer more momentum; the sweet spot gives the ball the most speed.
| Problem |
Setup |
Answer |
| 0.150 kg puck at 35.0 m/s caught by 70.0 kg goalie (Ex. 8.5a) |
v′ = 0.150 × 35.0 / 70.15 |
0.0748 m/s |
| KE lost (Ex. 8.5b) |
91.9 J before, 0.196 J after |
−91.7 J |
| Spring carts: 0.350 kg at 2.00 → −4.00 m/s; 0.500 kg at −0.500 m/s (Ex. 8.6) |
v₂′ = (0.700 − 0.250 + 1.40) / 0.500 |
3.70 m/s; KE gained 5.46 J from the spring |
Common mistakes: thinking KE is conserved in every collision; forgetting to add the masses when objects stick; sign errors when objects move toward each other.
Practice 8.5 in the quiz → · Top ↑
8.6 Collisions of Point Masses in Two Dimensions
In 2D, momentum is conserved separately along x and along y.
- Point mass — an object with no size that can't spin.
- Put m₂ at rest and m₁ moving along +x. Then the total y-momentum is zero before, so the y-momenta must cancel after: one object goes above the x-axis and the other below.
- Find each final velocity from its components: v = √(vₓ² + v_y²), θ = tan⁻¹(v_y / vₓ).
- Equal masses, elastic, target at rest: after a glancing collision the two move off 90° apart (billiard balls, without spin).
- Two momentum equations can solve for only two unknowns.
Worked example (Ex. 8.7): 0.250 kg at 2.00 m/s hits 0.400 kg at rest; afterward m₁ moves 1.50 m/s at 45.0°. Then tanθ₂ = (1.50 × 0.7071) / (1.50 × 0.7071 − 2.00) → θ₂ = 312° (48° below the x-axis) and v₂′ = 0.886 m/s. KE drops, so it's inelastic.
Practice 8.6 in the quiz → · Top ↑
8.7 Introduction to Rocket Propulsion
A rocket speeds up by throwing mass out the back; the exhaust pushes the rocket forward (Newton's third law).
- vₑ — exhaust speed relative to the rocket (about 2.5 × 10³ m/s for hot-gas rockets). Δm/Δt — fuel burn rate (kg/s).
- Acceleration goes up with faster exhaust, faster burning, and less mass. It increases during the flight as fuel is used up.
- The exhaust does not push on the ground or the air; rockets work better in a vacuum.
- Rocket equation (no gravity or drag, starting from rest): m₀ is the starting mass and m_r is what's left after the fuel burns.
- Multistage rockets drop empty tanks so later stages don't waste fuel moving dead weight.
| Problem |
Setup |
Answer |
| Moon rocket liftoff: m = 2.80×10⁶ kg, burn 1.40×10⁴ kg/s, vₑ = 2.40×10³ m/s (Ex. 8.8) |
Thrust = 3.36×10⁷ N; 3.36×10⁷ / 2.80×10⁶ = 12.0 m/s²; minus 9.80 |
2.20 m/s² |
| Mass ratio to reach 11.2 km/s with vₑ = 2.5 km/s |
ln(m₀/m_r) = 4.48 → m₀/m_r = e^4.48 |
88 (98.9% of the rocket is fuel) |
Common mistakes: forgetting to subtract g at liftoff; using the burn rate as a force; thinking rockets need air to push on.
Practice 8.7 in the quiz → · Top ↑
Chapter 11: Fluid Statics
11.1 What Is a Fluid?
Liquids, gases and plasmas are fluids because they flow when pushed sideways (shear); solids resist.
| Phase |
Shape |
Volume |
Compress easily? |
| Solid |
Definite |
Definite |
No |
| Liquid |
Takes container's |
Definite |
No |
| Gas |
Takes container's |
Fills container |
Yes |
| Plasma |
Takes container's |
Fills container |
Yes (charged particles) |
Practice 11.1 in the quiz → · Top ↑
11.2 Density
- Density (ρ) — mass per volume, in kg/m³. Water = 1000 kg/m³ = 1.000 g/cm³.
- Objects less dense than a fluid float in it; denser ones sink.
- Gases are about 1000× less dense than liquids and solids.
| Substance |
ρ (kg/m³) |
| Air |
1.29 |
| Ice |
917 |
| Water |
1000 |
| Seawater |
1025 |
| Blood |
1050 |
| Aluminum |
2700 |
| Iron / steel |
7800 |
| Mercury |
13,600 |
| Gold |
19,320 |
Ex. 11.1: a reservoir 50.0 km² × 40.0 m deep holds V = 2.00 × 10⁹ m³, so m = ρV = 2.00 × 10¹² kg.
Practice 11.2 in the quiz → · Top ↑
11.3 Pressure
- Pressure — force per area, perpendicular to the surface. Unit: pascal, 1 Pa = 1 N/m².
- Pressure is a scalar; in a fluid it pushes on every surface from every direction.
- Same force, smaller area means more pressure (a needle tip, cleats on turf).
Ex. 11.2: a tank at 6.90 × 10⁶ Pa with a 0.150 m diameter end: F = PA = 6.90 × 10⁶ × π(0.0750)² = 1.22 × 10⁵ N.
Practice 11.3 in the quiz → · Top ↑
11.4 Variation of Pressure with Depth
- Pressure grows with depth and depends only on depth and density, not on the shape or width of the container.
- 1 atm = 1.01 × 10⁵ Pa. Every 10.3 m of water adds about 1 atm.
- On a dam, use the average depth (half the full depth) to find the average pressure.
| Problem |
Setup |
Answer |
| Dam 500 m wide, 80.0 m deep (Ex. 11.3) |
P̄ = 40.0 × 1000 × 9.80 = 3.92 × 10⁵ Pa; F = P̄ × (80.0 × 500) |
1.57 × 10¹⁰ N |
| Depth of water for 1 atm (Ex. 11.5) |
h = 1.01 × 10⁵ / (1000 × 9.80) |
10.3 m |
Practice 11.4 in the quiz → · Top ↑
11.5 Pascal's Principle
A pressure change in an enclosed fluid is passed on undiminished to every part of it.
- Hydraulics multiply force: a big piston gets a big force from a small one. Car brakes, lifts, bulldozers.
- There's no free energy: the big piston moves a shorter distance (F₁d₁ = F₂d₂).
- Area goes with radius squared: 5× the diameter is 25× the area.
Ex. 11.6 (brakes): 500 N on a 0.500 cm cylinder drives 2.50 cm wheel cylinders: F₂ = (2.50/0.500)² × 500 = 1.25 × 10⁴ N each.
Practice 11.5 in the quiz → · Top ↑
11.6 Gauge Pressure, Absolute Pressure, and Measurement
- Gauge pressure — pressure above atmospheric (tire gauges, blood pressure). Can be negative.
- Absolute pressure — total pressure; never below zero.
- Manometers measure gauge pressure with a fluid column (P_g = hρg). Barometers measure atmospheric pressure (absolute).
- Mercury is 13.6× denser than water, so its columns are 13.6× shorter.
- Blood pressure: systolic/diastolic, typically 120/80 mm Hg.
| Unit |
Equals |
| 1 atm |
1.013 × 10⁵ Pa = 760 mm Hg = 14.7 psi = 10.3 m water |
| 1 mm Hg |
133 Pa |
| 1 psi |
6.90 × 10³ Pa |
| Problem |
Setup |
Answer |
| Tire gauge reads 34 psi |
P_abs = 34 + 14.7 |
48.7 psi |
| IV bag height for 18 mm Hg (Ex. 11.7) |
18 × 133 = 2400 Pa; h = 2400 / (1000 × 9.80) |
0.24 m |
Practice 11.6 in the quiz → · Top ↑
11.7 Archimedes' Principle
The buoyant force on an object equals the weight of the fluid it displaces.
- Buoyancy exists because pressure is higher at the bottom of an object than at the top.
- F_B > weight → rises and floats. F_B < weight → sinks (but still feels buoyancy). Equal → hovers.
- Specific gravity below 1 floats in water; above 1 sinks. For a floating object, the fraction submerged equals SG (ice: 0.917 in fresh water).
- Apparent weight in water = real weight − F_B. Weighing in air and in water gives density.
- Shape matters: steel shaped as a hull displaces enough water to float.
| Problem |
Setup |
Answer |
| Floating woman, 97.0% submerged (Ex. 11.9) |
ρ = 0.970 × 1000 |
970 kg/m³ |
| Coin: 8.630 g in air, 7.800 g in water (Ex. 11.10) |
V = 0.830 cm³; ρ = 8.630 / 0.830 |
10.4 g/cm³ (silver) |
| 1.00 × 10⁷ kg of steel as a solid block (Ex. 11.8) |
V = 1.28 × 10³ m³; F_B = 1.3 × 10⁷ N vs. weight 9.8 × 10⁷ N |
Sinks; as a hull displacing 1.00 × 10⁵ m³ it floats |
Practice 11.7 in the quiz → · Top ↑
11.8 Surface Tension and Capillary Action
- Cohesion — attraction between like molecules. Adhesion — between different ones.
- Surface tension (γ) — cohesion pulls a surface tight (insects walk on water). Water 0.0728 N/m, soapy water 0.037, mercury 0.465.
- Smaller bubbles have higher pressure inside. Lung surfactant keeps small alveoli from collapsing.
- Capillary action — water climbs a thin glass tube (contact angle 0°, adhesion wins); mercury drops (140°, cohesion wins). Narrower tube = higher rise.
| Problem |
Setup |
Answer |
| Soap bubble, r = 2.00 × 10⁻⁴ m (Ex. 11.11) |
P = 4 × 0.037 / 2.00 × 10⁻⁴ |
740 Pa (5.56 mm Hg) |
| Tube radius to lift sap 100 m (Ex. 11.12) |
r = 2(0.0728) / (1050 × 9.80 × 100) |
1.41 × 10⁻⁷ m (capillary action alone can't do it) |
Practice 11.8 in the quiz → · Top ↑
11.9 Pressures in the Body
- Body pressures are given as gauge pressures in mm Hg.
- Blood: ~120 systolic / ~80 diastolic at heart level. Standing adds about 108 mm Hg at the feet (1.4 m of blood), which is why blood pressure is taken on the upper arm.
- Eye 12–24 mm Hg (glaucoma is too high); chest cavity is negative, which keeps the lungs against the ribs; bladder 0–25 filling, 100–150 full.
- Spine and joints feel the biggest pressures (bad lifting can reach ~50 atm between vertebrae).
Ex. 11.13: an eardrum (1.00 cm²) tears at 3.00 N: P = 3.00 × 10⁴ Pa (226 mm Hg), reached at h = P/ρg = 3.06 m of water.
Practice 11.9 in the quiz → · Top ↑
Chapter 12: Fluid Dynamics
12.1 Flow Rate and Its Relation to Velocity
- Flow rate (Q) — volume per time (m³/s or L/s). Not the same as speed.
- Continuity — for a liquid, the same volume passes every point each second, so narrower means faster.
- For round pipes, A = πr², so v₂ = (r₁/r₂)² v₁.
- Branching: n₁A₁v₁ = n₂A₂v₂ (one aorta feeds billions of capillaries; blood slows there).
- 1 L = 10⁻³ m³.
| Problem |
Setup |
Answer |
| Hose r = 0.900 cm, Q = 0.500 L/s (Ex. 12.2a) |
v = 5.00 × 10⁻⁴ / π(0.00900)² |
1.96 m/s |
| Nozzle r = 0.250 cm (Ex. 12.2b) |
v = (0.900/0.250)² × 1.96 |
25.5 m/s |
| Aorta r = 10 mm, Q = 5.0 L/min (Ex. 12.3) |
v = Q / πr² |
0.27 m/s; about 5 × 10⁹ capillaries |
Practice 12.1 in the quiz → · Top ↑
12.2 Bernoulli's Equation
- It's conservation of energy per unit volume for a flowing fluid (no friction).
- Bernoulli's principle (same height): faster fluid has lower pressure.
- Examples: shower curtain bulges in, a car is pulled toward a passing truck, airplane wings (faster air over the top), sails, atomizers and paint sprayers (entrainment), pitot tubes measure air speed.
Ex. 12.4: water speeds from 1.96 to 25.5 m/s at the nozzle (P₂ = 1.01 × 10⁵ Pa): P₁ = P₂ + ½ρ(v₂² − v₁²) = 4.24 × 10⁵ Pa.
Practice 12.2 in the quiz → · Top ↑
12.3 General Applications of Bernoulli's Equation
- Torricelli's theorem — water leaving a hole h below the surface moves as fast as if it had fallen h.
- Power in a flow = (energy per volume) × (volume per second). A pump that only raises pressure supplies ΔP × Q.
Ex. 12.6: pump raises 40.0 L/s from 0.700 × 10⁶ to 1.62 × 10⁶ Pa: power = 0.920 × 10⁶ × 0.0400 = 36.8 kW (about 50 hp).
Practice 12.3 in the quiz → · Top ↑
12.4 Viscosity and Laminar Flow; Poiseuille's Law
- Viscosity (η) — fluid friction, in Pa·s. Water 1.0 × 10⁻³, blood 2.08 × 10⁻³, honey 2–10 Pa·s. Cold oil is more viscous.
- Laminar flow is smooth layers; turbulent flow has swirls and much more resistance.
- Resistance depends on r⁴: halve the radius and resistance goes up 16×. A 16% narrower artery (plaque) cuts flow in half.
Ex. 12.8: an IV needle (r = 0.150 mm, l = 2.50 cm, Q = 0.120 cm³/s, vein at 8.00 mm Hg) needs P₂ = 8ηlQ/πr⁴ + P₁ = 1.62 × 10⁴ Pa.
Practice 12.4 in the quiz → · Top ↑
12.5 The Onset of Turbulence
| Reynolds number (tube) |
Flow |
| Below about 2000 |
Laminar |
| 2000–3000 |
Unstable, can switch |
| Above about 3000 |
Turbulent |
- N_R has no units. Turbulence in arteries (plaque, bad heart valves) makes sounds a stethoscope can hear.
Ex. 12.9: the IV needle's flow (ρ = 1025, v = 1.70 m/s, r = 0.150 mm, η = 1.00 × 10⁻³) has N_R = 523, laminar.
Practice 12.5 in the quiz → · Top ↑
12.6 Motion of an Object in a Viscous Fluid
- Moving through still fluid is the same as standing in a moving fluid.
- N_R′ below about 1: laminar, drag ∝ v (Stokes' law for small spheres). Between 10 and 10⁶: drag ∝ v². Above 10⁶: fully turbulent.
- Terminal speed — drag plus buoyancy balance weight. Faster for dense, small objects in thin fluids. Skydivers spread out to fall slower; cyclists draft because drag goes with v².
Ex. 12.10: a 7.40 cm ball at 40.0 m/s in air: N_R′ = 1.29 × 40.0 × 0.0740 / 1.81 × 10⁻⁵ = 2.11 × 10⁵, so it has a turbulent wake.
Practice 12.6 in the quiz → · Top ↑
12.7 Diffusion, Osmosis and Related Processes
- Diffusion — molecules spread out by random motion, from high to low concentration. Distance grows with √t, so it's fast over micrometers and slow over centimeters.
- Diffusion constant D — bigger for light molecules, gases, and higher temperatures. O₂ in air 1.8 × 10⁻⁵ m²/s; O₂ in water 1.0 × 10⁻⁹; glucose in water 6.7 × 10⁻¹⁰.
- Osmosis — water moves through a semipermeable membrane toward the side with more dissolved stuff (less water). Dialysis — other molecules doing the same.
- Osmotic pressure — back pressure that stops osmosis. Reverse osmosis pushes harder to filter (desalination).
- Active transport — cells spend energy to move molecules against the concentration difference.
Ex. 12.11: glucose spreading 1.0 cm in water: t = x²/2D = (0.010)² / (2 × 6.7 × 10⁻¹⁰) = 7.5 × 10⁴ s (21 h).
Common mistakes (Unit 6): using gauge pressure where absolute is needed (or the reverse); diameter instead of radius in A = πr²; forgetting r⁴ in Poiseuille's law; using the object's volume instead of the submerged volume for buoyancy; mixing L and m³.
Practice 12.7 in the quiz → · Top ↑
Chapter 13: Temperature, Kinetic Theory, and the Gas Laws
13.1 Temperature
Temperature is what a thermometer measures. Later in the chapter it gets a deeper meaning: the average kinetic energy of the molecules.
- Any property that changes reliably with temperature can make a thermometer: liquid expanding (alcohol), a bimetallic strip bending, electrical resistance, color (liquid crystal), infrared light (ear thermometers).
- Your hand is a bad thermometer. A metal bench feels colder than a wooden one at the same temperature because metal pulls heat out of your skin faster.
- Kelvin is the absolute scale: 0 K is absolute zero (−273.15 °C), the lowest possible temperature. Write "K", not "°K". Gas pressures at constant volume all extrapolate to zero at this same temperature.
- A change of 1 K equals a change of 1 °C. A change of 1 °C is 1.8× a change of 1 °F (100 C° span the same range as 180 F°).
- Thermal equilibrium — two objects in contact reach the same temperature, and heat stops flowing. A thermometer actually reads its own temperature once it's in equilibrium with the object.
- Zeroth law of thermodynamics — if A is in equilibrium with B, and B is in equilibrium with C, then A is in equilibrium with C. That's what makes temperature a meaningful thing to measure.
| Reference point |
°F |
°C |
K |
| Absolute zero |
−459.67 |
−273.15 |
0 |
| Water freezes |
32 |
0 |
273.15 |
| Room temperature |
77 |
25 |
298 |
| Body temperature |
98.6 |
37.0 |
310 |
| Water boils |
212 |
100 |
373.15 |
Ex. 13.1: room temperature 25 °C: T = (9/5)(25) + 32 = 77 °F, and T = 25 + 273.15 = 298 K.
Ex. 13.2 (Reaumur scale, water freezes at 0 °R and boils at 80 °R): T = 0.8 × 25 = 20 °R.
Practice 13.1 in the quiz → · Top ↑
13.2 Thermal Expansion of Solids and Liquids
Hotter atoms vibrate harder and push their neighbors a little farther away, so the whole object grows.
- α — coefficient of linear expansion, in 1/°C (same as 1/K, since the degree sizes match). β — coefficient of volume expansion.
- Holes get bigger when heated, just as if the plug were still there.
- Liquids expand much more than solids, so a full gas tank overflows on a hot day and a thermometer's alcohol rises in the glass.
- Water is the exception: between 0 °C and 4 °C it shrinks as it warms. Water is densest at 4 °C, so ponds freeze from the top down and fish survive in 4 °C water below the ice.
- Thermal stress — if something can't expand, it pushes back hard. Pressure = (ΔV/V₀) × B, where B is the bulk modulus. That's why bridges and train tracks need expansion joints and why Pyrex (small α) survives fast cooling.
| Material |
α (1/°C) |
β (1/°C) |
| Aluminum |
25 × 10⁻⁶ |
75 × 10⁻⁶ |
| Brass |
19 × 10⁻⁶ |
56 × 10⁻⁶ |
| Copper |
17 × 10⁻⁶ |
51 × 10⁻⁶ |
| Iron or steel |
12 × 10⁻⁶ |
35 × 10⁻⁶ |
| Concrete |
12 × 10⁻⁶ |
36 × 10⁻⁶ |
| Glass (ordinary) |
9 × 10⁻⁶ |
27 × 10⁻⁶ |
| Glass (Pyrex) |
3 × 10⁻⁶ |
9 × 10⁻⁶ |
| Water |
— |
210 × 10⁻⁶ |
| Mercury |
— |
180 × 10⁻⁶ |
| Gasoline (petrol) |
— |
950 × 10⁻⁶ |
| Air (and most gases) |
— |
3400 × 10⁻⁶ |
| Problem |
Setup |
Answer |
| Golden Gate Bridge, 1275 m of steel, −15 °C to 40 °C (Ex. 13.3) |
ΔL = 12 × 10⁻⁶ × 1275 × 55 |
0.84 m |
| 60.0 L steel tank full of gas, 15.0 °C → 35.0 °C (Ex. 13.4) |
V_spill = (950 − 35) × 10⁻⁶ × 60.0 × 20.0 |
1.10 L spills |
| Same gas, not allowed to expand, B = 1.00 × 10⁹ Pa (Ex. 13.5) |
P = (1.10 / 60.0) × 1.00 × 10⁹ |
1.83 × 10⁷ Pa (about 2500 psi) |
Practice 13.2 in the quiz → · Top ↑
13.3 The Ideal Gas Law
Gas molecules are far apart compared with their size, so all gases act almost the same way. An ideal gas has tiny molecules that only interact by bouncing (elastic collisions).
- P must be absolute pressure (not gauge) and T must be in kelvin. This is the #1 mistake in this chapter.
- N = number of molecules. n = number of moles. Pick the form of the law that matches what you know.
- Avogadro's number: 1 mole = 6.02 × 10²³ particles. A mole has a mass in grams equal to the molecular mass (water 18.0 g/mol, air 28.8 g/mol).
- Boyle's law (fixed T): PV = constant. Charles' law (fixed P): V ∝ T.
- For a fixed amount of gas, take a ratio of final to initial states so the unknowns cancel (e.g. P_f = P₀ × T_f/T₀ at constant V).
- At STP (0 °C, 1 atm), 1 mole of any gas fills 22.4 L.
- PV has units of joules: a gas stores energy in its pressure and volume.
| Constant |
Value |
| Boltzmann constant k |
1.38 × 10⁻²³ J/K |
| Avogadro's number N_A |
6.02 × 10²³ mol⁻¹ |
| Gas constant R = N_A k |
8.31 J/(mol·K) = 0.0821 L·atm/(mol·K) = 1.99 cal/(mol·K) |
| 1 atm |
1.01 × 10⁵ Pa |
| Problem |
Setup |
Answer |
| Bike tire at 7.00 × 10⁵ Pa, 18.0 °C → 35.0 °C (Ex. 13.6) |
P_f = 7.00 × 10⁵ × 308/291 |
7.41 × 10⁵ Pa |
| Molecules in 1.00 m³ at STP (Ex. 13.7) |
N = (1.01 × 10⁵)(1.00) / (1.38 × 10⁻²³ × 273) |
2.68 × 10²⁵ molecules |
| Moles per m³ and liters per mole at STP (Ex. 13.8) |
2.68 × 10²⁵ / 6.02 × 10²³; 1000 / 44.5 |
44.5 mol/m³; 22.5 L/mol |
| Moles in a 2.00 L tire at 7.00 × 10⁵ Pa, 18.0 °C (Ex. 13.9) |
n = (7.00 × 10⁵)(2.00 × 10⁻³) / (8.31 × 291) |
0.579 mol |
Practice 13.3 in the quiz → · Top ↑
13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
Kinetic theory — gas pressure comes from huge numbers of molecules bouncing off the walls. Each bounce flips the molecule's momentum (Δp = 2mv), which pushes on the wall.
- Average kinetic energy depends only on absolute temperature, not on which gas it is. This is the molecular meaning of temperature. The average translational KE of a molecule is called its thermal energy.
- Double the kelvin temperature → double the average KE → v_rms goes up by √2 (not 2).
- Lighter molecules move faster at the same temperature (helium beats nitrogen), which is why helium leaks out of Earth's atmosphere.
- Mass of one molecule: m = (molar mass in kg/mol) / N_A. N₂ is 28.0 g/mol = 0.0280 kg/mol.
- Maxwell-Boltzmann distribution — molecules have a spread of speeds with a long tail toward high speeds. The most probable speed v_p is less than v_rms. Higher temperature shifts the curve to faster speeds and makes it wider.
- Molecules move fast (about 500 m/s) but collide constantly, so they don't get far. The speed of sound (about 340 m/s) is tied to these speeds.
| Problem |
Setup |
Answer |
| Average KE of a gas molecule at 20.0 °C (Ex. 13.10a) |
KE = 1.5 × 1.38 × 10⁻²³ × 293 |
6.07 × 10⁻²¹ J |
| v_rms of N₂ at 20.0 °C, m = 4.65 × 10⁻²⁶ kg (Ex. 13.10b) |
v = √(3 × 1.38 × 10⁻²³ × 293 / 4.65 × 10⁻²⁶) |
511 m/s |
| T for helium (m = 6.65 × 10⁻²⁷ kg) to reach v_rms = 11.1 km/s (Ex. 13.11) |
T = m v² / 3k |
1.98 × 10⁴ K |
Practice 13.4 in the quiz → · Top ↑
13.5 Phase Changes
At low temperatures or high pressures, real gases stop acting ideal and turn into liquids and solids.
- PV diagram — curves of P vs. V at a fixed temperature are isotherms (hyperbolas for an ideal gas).
- Critical point — above the critical temperature, a substance can't be liquid at any pressure. CO₂: 31.1 °C. Water: 374 °C at 219 atm. The critical pressure is the minimum pressure for liquid to exist at the critical temperature.
- Phase diagram — a graph of P vs. T showing solid, liquid and gas regions. On a boundary line, two phases coexist.
- Higher pressure raises water's boiling point (pressure cooker cooks faster). Higher pressure lowers ice's melting point (snowballs, tires on snow).
- Sublimation — solid straight to gas. Water does this below 0.00600 atm (freeze drying, freezer defrost). Dry ice (CO₂) sublimates at 1 atm.
- Triple point — the one P and T where solid, liquid and gas all coexist. Water: 273.16 K (0.01 °C) at 0.00600 atm.
- Equilibrium — at the boiling point in a closed container, molecules evaporate and condense at the same rate. An open pot boils away because air, not pure steam, sits above it.
- Vapor — the gas phase below its boiling temperature. Vapor pressure — the pressure at which the gas coexists with its liquid or solid; it rises with temperature.
- Partial pressure — the pressure one gas in a mixture would make alone. Dalton's law: total pressure = sum of the partial pressures.
- Water evaporates when its vapor pressure is higher than the partial pressure of water vapor in the air; dew or frost forms when it's lower.
| Substance |
Critical T (°C) |
Critical P (atm) |
Triple point T (K) |
Triple point P (atm) |
| Water |
374.3 |
219.0 |
273.16 |
0.00600 |
| Carbon dioxide |
31.1 |
73.2 |
216.55 |
5.11 |
| Nitrogen |
−146.9 |
33.6 |
63.18 |
0.124 |
| Oxygen |
−118.4 |
50.3 |
54.36 |
0.00151 |
Practice 13.5 in the quiz → · Top ↑
13.6 Humidity, Evaporation, and Boiling
"It's not the heat, it's the humidity." Sweat cools you by evaporating, and humid air slows evaporation.
- Relative humidity compares the water vapor in the air to the most possible at that temperature (saturation, 100%).
- Air doesn't really "hold" water. The amount of vapor is set by water's vapor pressure, which rises with temperature. Warmer air can have more vapor.
- Dew point — the temperature where the current vapor density becomes the saturation density (100% humidity). Cool below it and dew or fog forms. Low dew point = dry air (deserts swing hot and cold).
- Boiling happens when water's vapor pressure equals the surrounding pressure (1.01 × 10⁵ Pa at 100 °C). Bubbles of vapor grow, buoyancy lifts them, and they rise.
- Use the ideal gas law to turn a vapor pressure into a vapor density: n/V = P/RT, then multiply by 18.0 g/mol.
| T (°C) |
Vapor pressure (Pa) |
Saturation vapor density (g/m³) |
| −10 |
2.60 × 10² |
2.36 |
| 0 |
6.10 × 10² |
4.84 |
| 10 |
1.19 × 10³ |
9.40 |
| 15 |
1.69 × 10³ |
12.8 |
| 20 |
2.33 × 10³ |
17.2 |
| 25 |
3.17 × 10³ |
23.0 |
| 30 |
4.24 × 10³ |
30.4 |
| 37 |
6.31 × 10³ |
44.0 |
| 40 |
7.34 × 10³ |
51.1 |
| 100 |
1.01 × 10⁵ |
598 |
| Problem |
Setup |
Answer |
| Vapor density at 20.0 °C from P = 2.33 × 10³ Pa (Ex. 13.12) |
n/V = 2.33 × 10³ / (8.31 × 293) = 0.957 mol/m³; × 18.0 |
17.2 g/m³ (matches the table) |
| 25.0 °C air with 9.40 g/m³ (Ex. 13.13a, b) |
9.40 / 23.0 × 100; dew point where saturation = 9.40 |
40.9%; dew point 10.0 °C |
| 25.0 °C, dew point −10.0 °C (Ex. 13.13c) |
2.36 / 23.0 × 100 |
10.3% |
Common mistakes (Chapter 13): using °C instead of K in the gas law or in KE = (3/2)kT (ΔT is fine in either); using gauge pressure instead of absolute pressure; mixing up N (molecules) with n (moles), or k with R; using grams per mole instead of kg per mole when finding a molecule's mass; forgetting the square root in v_rms; using α instead of β for volume expansion; thinking air "holds" water vapor.
Practice 13.6 in the quiz → · Top ↑
Chapter 14: Heat and Heat Transfer Methods
14.1 Heat
Heat is energy that moves on its own because of a temperature difference. It always flows from hot to cold and stops at thermal equilibrium (same temperature).
- Heat is energy, so its SI unit is the joule (J). Temperature is not energy; it measures the average kinetic energy of the molecules.
- Calorie (cal) — energy to warm 1.00 g of water by 1.00 °C. Kilocalorie (kcal) — 1.00 kg of water by 1.00 °C.
- A food Calorie (capital C) is really a kilocalorie.
- Mechanical equivalent of heat — Joule showed that work (a falling weight stirring water) warms water just like heating it. Work and heat both change a system's internal energy.
- A system has internal energy, not "heat content." Heat is energy in transit.
- Adding heat doesn't always raise temperature: during melting or boiling it doesn't (see 14.3).
Common mistake: mixing kcal and J. Multiply kcal by 4186 to get J; a 250 Cal snack is 250 kcal = 1.05 × 10⁶ J.
Practice 14.1 in the quiz → · Top ↑
14.2 Temperature Change and Heat Capacity
- Heat needed depends on mass, temperature change, and the substance and phase (its specific heat c).
- Specific heat (c) — heat to raise 1.00 kg by 1.00 °C, in J/(kg·°C). A ΔT of 1 °C equals a ΔT of 1 K, so J/(kg·K) is the same unit.
- Water's c is huge: 5× glass, about 10× iron. That's why lakes and oceans change temperature slowly and keep coastal weather mild.
- Calorimetry — two things in contact trade heat until they reach the same final temperature. Heat lost by the hot one = heat gained by the cold one (if no heat escapes).
- Work can do the same job as heating: friction in truck brakes turns gravitational PE into internal energy.
| Substance |
c (J/kg·°C) |
| Water (15 °C) |
4186 |
| Human body (avg) |
3500 |
| Ethanol |
2450 |
| Ice (−50 to 0 °C) |
2090 |
| Steam (100 °C, constant P) |
2020 |
| Wood |
1700 |
| Air (dry, constant P) |
1015 |
| Aluminum |
900 |
| Glass, concrete |
840 |
| Iron, steel |
452 |
| Copper |
387 |
| Silver |
235 |
| Mercury |
139 |
| Gold |
129 |
| Lead |
128 |
Ex. 14.1: 0.500 kg aluminum pan + 0.250 kg water, heated from 20.0 to 80.0 °C: Q_w = 0.250 × 4186 × 60.0 = 62.8 kJ, Q_Al = 0.500 × 900 × 60.0 = 27.0 kJ → 89.8 kJ total (69.9% to the water).
Ex. 14.2: a 10,000 kg truck drops 75.0 m; 10% of Mgh = 7.35 × 10⁵ J goes into 100 kg of brake material (c = 800): ΔT = 7.35 × 10⁵ / (100 × 800) → 9.2 °C.
Ex. 14.3: 0.250 kg of 20.0 °C water poured into a 0.500 kg aluminum pan at 150 °C: T_f = (0.500·900·150 + 0.250·4186·20.0) / (0.500·900 + 0.250·4186) → 59.1 °C (close to the water's start, because water's c is big).
Common mistakes: using the final temperature instead of ΔT; forgetting the container (the pan) also soaks up heat; using liters as kilograms for anything but water.
Practice 14.2 in the quiz → · Top ↑
14.3 Phase Change and Latent Heat
- During a phase change the temperature stays constant until the change is finished. The energy breaks (or forms) bonds between molecules instead of speeding them up.
- Latent heat of fusion (L_f) — energy per kg to melt. Latent heat of vaporization (L_v) — energy per kg to boil. Unit: J/kg. "Latent" means hidden: no temperature change shows it.
- Freezing and condensing release the same energy that melting and boiling absorb. (Fruit growers spray water on oranges; the freezing water gives off heat and protects the fruit.)
- Phase-change energies are big: melting 1 kg of ice (334 kJ) takes as much energy as warming that water from 0 to 79.8 °C. Boiling it takes 2256 kJ, almost 7× more.
- Evaporation happens below boiling too and takes even more energy per kg (2430 kJ/kg at 37 °C). Sweat cools you this way; humid air slows it.
- Sublimation — solid straight to vapor (dry ice, snow vanishing on cold sunny days). The reverse is frost forming on a window.
- Heating curve for water: warm the ice (c = 2090) → flat plateau at 0 °C (melting) → warm the water (c = 4186) → flat plateau at 100 °C (boiling) → warm the steam (c = 2020).
| Substance |
Melting pt (°C) |
L_f (kJ/kg) |
Boiling pt (°C) |
L_v (kJ/kg) |
| Water |
0.00 |
334 |
100.0 |
2256 (2430 at 37 °C) |
| Ethanol |
−114 |
104 |
78.3 |
854 |
| Nitrogen |
−210.0 |
25.5 |
−195.8 |
201 |
| Mercury |
−38.9 |
11.8 |
357 |
272 |
| Lead |
327 |
24.5 |
1750 |
871 |
| Aluminum |
660 |
380 |
2450 |
11,400 |
| Copper |
1083 |
134 |
2595 |
5069 |
| Gold |
1063 |
64.5 |
2660 |
1578 |
Ex. 14.4: three 6.0 g ice cubes at 0 °C chill 0.25 kg of soda at 20 °C. Ice first melts (0.018 × 334,000 = 6012 J), then that meltwater warms up: T_f = (20,930 − 6012) / 1122 → 13 °C.
Problem strategy: check whether a melting or boiling point is crossed. If so, work in steps: temperature change → phase change → temperature change, then add the Q's.
Common mistakes: skipping the plateau (forgetting mL_f or mL_v); using water's c for ice or steam; using kJ/kg values as J/kg (334 kJ/kg = 334,000 J/kg).
Practice 14.3 in the quiz → · Top ↑
14.4 Heat Transfer Methods
Every heat transfer happens by one of three methods, and all three are driven only by a temperature difference.
| Method |
How it works |
Example |
| Conduction |
Collisions between molecules in matter that isn't flowing; needs contact |
Stove burner → bottom of a pan; metal bat handle on a cold night |
| Convection |
A fluid (liquid or gas) moves and carries the energy with it |
Forced-air furnace, weather, boiling pot, car cooling system |
| Radiation |
Electromagnetic waves (mostly infrared) are emitted and absorbed; no medium needed |
Sun warming Earth, feeling a campfire from across the yard |
- A fireplace uses all three; most of the heat reaching the room comes by radiation.
- We control heat flow by choice of material (wool, foam), stopping air movement (weather stripping), and color (a white roof).
Practice 14.4 in the quiz → · Top ↑
14.5 Conduction
- Rate of conduction (in watts, J/s) grows with conductivity k, area A, and temperature difference, and drops with thickness d.
- Fast molecules bump slower ones and pass on energy. More area = more collisions; thicker = longer path.
- Good electrical conductors (silver, copper, gold, aluminum) are good heat conductors because of their free electrons. Pans are metal.
- Good insulators (Styrofoam, down, wool, fiberglass) trap tiny pockets of air, which conducts poorly.
- R factor = d/k — bigger R means better insulation. US values are in ft²·°F·h/Btu: R-11 for 3.5 in fiberglass, R-19 for 6.5 in.
- Tile feels colder than carpet at the same temperature because tile conducts heat out of your feet faster.
| Substance |
k (J/s·m·°C) |
| Silver |
420 |
| Copper |
390 |
| Gold |
318 |
| Aluminum |
220 |
| Steel (iron) |
80 |
| Stainless steel |
14 |
| Ice |
2.2 |
| Glass, concrete brick |
0.84 |
| Water |
0.6 |
| Fatty tissue |
0.2 |
| Wood |
0.08–0.16 |
| Snow (dry) |
0.10 |
| Wool |
0.04 |
| Down feathers |
0.025 |
| Air |
0.023 |
| Styrofoam |
0.010 |
Ex. 14.5: a Styrofoam cooler (A = 0.950 m², d = 2.50 cm) at 0 °C inside, 35.0 °C outside: Q/t = 0.010 × 0.950 × 35.0 / 0.0250 = 13.3 W; in one day Q = 1.15 × 10⁶ J, melting m = Q/L_f → 3.44 kg of ice.
Ex. 14.6: water boils off at 1.00 g/s (2256 W) through an aluminum pan bottom 0.800 cm thick and 14.0 cm across: ΔT = (Q/t)·d/(kA) = 2256 × 0.00800 / (220 × 0.0154) → 5.33 °C.
Common mistakes: leaving d in cm (convert to m); using diameter for the radius when finding A; forgetting Q/t is a rate, so multiply by time (in seconds) to get Q.
Practice 14.5 in the quiz → · Top ↑
14.6 Convection
- Convection — heat carried by a moving fluid. Natural convection is driven by buoyancy: hot fluid expands, gets less dense, and rises; cool fluid sinks. Forced convection uses a fan or pump.
- Examples: a pot of water rolling on the stove, room heating loops, ocean currents and winds, the car's water pump, blood flow to the skin when you're hot.
- To calculate it, find the mass of fluid moved and use Q = mcΔT (or mL for a phase change), then divide by time.
- Wind chill — moving air strips heat much faster than still air. A 15 m/s wind at 0 °C feels like still air at about −18 °C.
- Air is a poor conductor, but only if it can't move. Small gaps (double-pane windows, fur, insulation) stop convection loops, so the trapped air insulates.
- Convection + phase change: sweat evaporating needs moving air to carry away the humid air, so a fan feels cool. Water vapor rising and condensing releases huge energy that powers thunderheads and hurricanes (positive feedback).
Ex. 14.7: a 12.0 × 18.0 × 3.00 m house (m = 1.29 × 648 = 836 kg of air, c ≈ 1000) has its air replaced every 30.0 min, warmed by 10.0 °C: Q/t = 836 × 1000 × 10.0 / 1800 → 4.64 kW.
Ex. 14.8: a resting person makes 120 W. To shed it all by sweating (L_v = 2430 J/g at body temperature): m/t = 120 / 2430 = 0.0494 g/s ≈ 2.96 g/min.
Practice 14.6 in the quiz → · Top ↑
14.7 Radiation
- Radiation — energy carried by electromagnetic waves (infrared, visible, ...). It needs no medium, so it crosses empty space from the Sun.
- Every object above absolute zero emits and absorbs radiation. Hotter objects radiate far more and glow red → orange → yellow → white (→ blue).
- Emissivity (e) — how well a surface radiates, from 0 (perfect mirror) to 1 (ideal blackbody). Good absorbers are good emitters. Black asphalt gets hotter in the sun and colder on a clear night than gray concrete.
- Rate ∝ T⁴ with T in kelvin. Doubling the kelvin temperature makes it 16× bigger.
- Rate ∝ area: spreading out fire coals makes them radiate more.
- Net radiation uses T₂ = surroundings and T₁ = object. Negative means the object is losing heat.
- Greenhouse effect — the atmosphere lets visible sunlight in, but CO₂ and H₂O absorb Earth's longer-wavelength infrared and send some back. It keeps Earth about 40 °C warmer than it would be. More CO₂ traps more.
- Earth's average e is about 0.65; clouds reflect and change it (a negative feedback).
| Surface |
Emissivity e |
| Ideal blackbody |
1 |
| Carbon black |
0.99 |
| Human skin (infrared) |
0.97 |
| Earth (average) |
0.65 |
| Tungsten filament |
about 0.5 |
| Perfect reflector |
0 |
Ex. 14.9: unclothed person, skin 33.0 °C (306 K), room 22.0 °C (295 K), A = 1.50 m², e = 0.97: Q/t = 5.67 × 10⁻⁸ × 0.97 × 1.50 × (295⁴ − 306⁴) → −99 W (losing about 99 W).
Common mistakes: plugging °C into T⁴ (always add 273); computing (T₂ − T₁)⁴ instead of T₂⁴ − T₁⁴; forgetting e.
Common mistakes (Chapter 14): kcal vs J (× 4186); skipping a phase-change plateau; using water's c for ice or steam; ΔT is the same in K and °C but T⁴ needs kelvin; cm instead of m in conduction thickness; forgetting Q/t is a rate (watts).
Practice 14.7 in the quiz → · Top ↑
Chapter 15: Thermodynamics
15.1 The First Law of Thermodynamics
The first law is conservation of energy for a system where energy moves in or out by heat and work.
- Internal energy (E_int) — the total kinetic plus potential energy of all the atoms and molecules. It depends only on the state (P, V, T), not on how the system got there. ΔE_int is path independent.
- Heat (Q) — energy moved because of a temperature difference. Random, disorganized. Depends on the path.
- Work (W) — energy moved by a force acting through a distance. Organized. Depends on the path.
- Heat and work are energy in transit. A system doesn't "contain" heat or work; it contains internal energy.
- Once a gas warms up, you can't tell whether heat or work did it (a bike tire warms in the Sun or when you pump it).
| Quantity |
Positive (+) means |
Negative (−) means |
| Q |
Heat flows into the system |
Heat flows out of the system |
| W |
Work done by the system (gas expands) |
Work done on the system (gas is compressed) |
| ΔE_int |
Internal energy goes up |
Internal energy goes down |
Ex. 15.1a: 40.00 J of heat in and 10.00 J of work done by the system, then 25.00 J of heat out and 4.00 J of work done on it: Q = 15.00 J, W = 6.00 J, so ΔE_int = 9.00 J.
Ex. 15.1b: 150.00 J of heat out (Q = −150.00 J) and 159.00 J of work done on the system (W = −159.00 J): ΔE_int = −150.00 − (−159.00) = 9.00 J. Different path, same change.
Human metabolism
- The body loses heat (Q < 0) and does work (W > 0), so it loses internal energy. Food puts it back (count food intake as work done on the body).
- Eat more than you use: ΔE_int > 0 and the extra is stored as fat. Eat less: the body burns fat.
- 1 food Calorie = 1 kcal = 4186 J (enough to warm 1 kg of water by 1 °C). A lowercase calorie is 1000 times smaller.
- Basal metabolic rate (BMR) — the rate the body uses energy at complete rest. It drops when you diet, which slows weight loss.
- Many processes are irreversible: fat can become work and heat, but sunbathing can't make fat.
Practice 15.1 in the quiz → · Top ↑
15.2 The First Law of Thermodynamics and Some Simple Processes
- A heat engine uses heat transfer to do work (car engines, steam turbines). Some heat always goes out to the environment (Q_out is never 0).
- On a PV diagram, the work done is the area under the curve. Expanding (ΔV > 0) gives positive W; compressing gives negative W.
- Work depends on the path: the higher-pressure path between the same two points does more work.
- For a cycle (returns to start), net work = area inside the loop. Clockwise loop: net work out (engine). Counterclockwise: net work in (fridge/heat pump).
| Process |
What stays constant |
Key fact |
First law becomes |
| Isobaric |
Pressure |
W = PΔV |
ΔE_int = Q − PΔV |
| Isochoric |
Volume |
W = 0 (vertical line on PV) |
ΔE_int = Q |
| Isothermal |
Temperature |
ΔE_int = 0 (ideal gas) |
Q = W |
| Adiabatic |
No heat flow |
Q = 0 |
ΔE_int = −W |
- Monatomic ideal gas: E_int = (3/2)NkT = (3/2)nRT, so internal energy only changes if temperature changes.
- Adiabatic expansion cools the gas (work comes out of internal energy). That's why gas from a spray can or CO₂ cartridge feels cold. Adiabatic happens with good insulation or a very fast process.
- Isothermal processes are slow (heat must keep flowing in). From the same start, an isothermal expansion does more work than an adiabatic one.
- Reversible process — both the system and its surroundings can return exactly to where they started. Real processes are never fully reversible (friction, turbulence).
Ex. 15.2: a rectangle cycle ABCDA with P_AB = 1.50 × 10⁶ Pa, P_CD = 2.00 × 10⁵ Pa, ΔV = 5.00 × 10⁻⁴ m³: W_AB = 750 J, W_CD = −100 J, the two vertical legs do 0, so W = (1.50 × 10⁶ − 2.00 × 10⁵)(5.00 × 10⁻⁴) = 650 J.
Practice 15.2 in the quiz → · Top ↑
15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
Second law (first form): heat flows on its own from hot to cold, never from cold to hot.
Second law (second form): in a cycle, no engine can turn heat from a reservoir completely into work. Some Q_c must go out.
- The first law would allow a cold object to get colder while warming a hot one; the second law forbids it. It sets the direction of natural processes.
- Q_h = heat in from the hot reservoir, Q_c = heat out to the cold reservoir, W = work out. In these formulas all three are positive numbers.
- For a full cycle ΔE_int = 0, so W = Q_h − Q_c.
- Efficiency is always less than 1 (less than 100%). Coal plants are about 42%; nuclear about 35%; car engines much lower.
- CO₂ from coal: C + O₂ → CO₂, so 12 kg of carbon makes 44 kg of CO₂.
Ex. 15.3: a coal plant takes in Q_h = 2.50 × 10¹⁴ J per day and dumps Q_c = 1.48 × 10¹⁴ J. W = 1.02 × 10¹⁴ J; Eff = 1.02/2.50 = 0.408 (40.8%). At 2.5 × 10⁶ J/kg it burns 1.0 × 10⁸ kg of coal, making 1.0 × 10⁸ × 44/12 = 3.7 × 10⁸ kg of CO₂ per day.
The Otto cycle (gasoline engine)
| Path |
Process |
Engine stroke |
| A→B |
Adiabatic compression (work done on gas) |
Compression stroke |
| B→C |
Isochoric, Q_h added |
Spark ignites fuel |
| C→D |
Adiabatic expansion (gas does work) |
Power stroke |
| D→A |
Isochoric, Q_c released |
Exhaust + intake |
- Four strokes: intake, compression, power, exhaust. Net work = area inside the loop.
- Hotter power stroke and cooler compression make a bigger loop and a more efficient engine.
Practice 15.3 in the quiz → · Top ↑
15.4 Carnot's Perfect Heat Engine: The Second Law of Thermodynamics Restated
- Carnot cycle — two isothermal plus two adiabatic steps, all reversible. It's the most efficient cycle possible.
- Second law (third form): a Carnot engine has the greatest possible efficiency of any engine between two temperatures. Every reversible engine between those temperatures has that same efficiency.
- For a Carnot engine, Q_c/Q_h = T_c/T_h.
- Temperatures must be in kelvin. Using °C gives nonsense.
- 100% efficiency would need T_c = 0 K, which is impossible.
- Best efficiency: hot reservoir as hot as possible, cold reservoir as cold as possible.
- Real engines are less efficient than Carnot because of dissipative (irreversible) processes such as friction and turbulence, which send more energy to Q_c. A Carnot engine would also have zero power (it's infinitely slow).
Ex. 15.4: a nuclear reactor with water at 300 °C (573 K) and a condenser at 27 °C (300 K): Eff_C = 1 − 300/573 = 0.476 (47.6%). The real plant gets about 35%.
Practice 15.4 in the quiz → · Top ↑
15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
Heat pumps, air conditioners and refrigerators are heat engines run backward: work input W moves heat from cold to hot.
- Heat pump — its job is to put Q_h into a warm space (your house in winter). You pay only for W; Q_c comes free from the outside air, even below freezing.
- Refrigerator / AC — its job is to pull Q_c out of a cold space. Its waste heat is the heat pump's "product."
- Coefficient of performance (COP) — what you get ÷ what you pay. COP_hp is always greater than 1. Real heat pumps are 2–4; real fridges and ACs 2–6.
- Best (Carnot) COP: COP_hp = 1 / Eff_C = T_h / (T_h − T_c). Heat pumps work best when the temperature difference is small, so they struggle in very cold climates.
- Parts of a heat pump: evaporator (outdoor coils take in Q_c), compressor (work W), condenser (indoor coils give off Q_h), expansion valve.
- EER (energy efficiency rating) uses mixed units, Btu per hour ÷ watts. Higher EER = cheaper to run.
Ex. 15.5: a heat pump with T_h = 45.0 °C (318 K) and T_c = −15.0 °C (258 K): Eff_C = 1 − 258/318 = 0.1887, so COP_hp = 1/0.1887 = 5.30. Each 1 J of electricity delivers 5.30 J to the house (4.3 J of it from the cold outside air).
Practice 15.5 in the quiz → · Top ↑
15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
- Entropy (S) — a measure of disorder and of energy unavailable to do work. Unit: J/K.
- Q is + for heat in (entropy goes up) and − for heat out (entropy goes down). T is the absolute temperature (use the average if T changes a little).
- Entropy is a state property, like E_int, so you can find ΔS for a real process by imagining a reversible path between the same states.
- Reversible process: total ΔS = 0. Irreversible (every real process): total ΔS > 0.
- Second law (fourth form): the total entropy of a system increases or stays the same; it never decreases.
- Heat flowing hot → cold raises total entropy because the same Q is divided by a smaller T on the cold side.
- Entropy is not conserved (energy is).
- W_unavail = ΔS · T₀, where T₀ is the lowest temperature available.
- Heat death of the universe — eventually all temperatures even out, entropy is maximum, and no work is possible.
- Life and evolution don't break the second law. Entropy can drop locally (a plant, a bridge) as long as the environment's entropy rises more: ΔS_tot = ΔS_syst + ΔS_envir > 0. Earth is a heat engine between the hot Sun and cold space.
| Problem |
Setup |
Answer |
| 4000 J flows from 600 K to 250 K (Ex. 15.6) |
ΔS_h = −4000/600 = −6.67 J/K; ΔS_c = 4000/250 = 16.0 J/K |
ΔS_tot = 9.33 J/K |
| Carnot engine 600 K → 100 K with 4000 J (Ex. 15.7a) |
Eff = 1 − 100/600 = 0.833 |
W = 3333 J |
| Same 4000 J first drops to 250 K (Ex. 15.7b) |
Eff = 1 − 100/250 = 0.600 |
W = 2400 J; 933 J lost = 9.33 × 100 |
| Melt 1.00 kg of ice at 0 °C (Ex. 15.8) |
Q = mL_f = 3.34 × 10⁵ J; ΔS = Q/273 |
1.22 × 10³ J/K |
Practice 15.6 in the quiz → · Top ↑
15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
The second law happens because disorder is overwhelmingly more likely than order.
- Macrostate — the overall result (3 heads, 2 tails). Microstate — one exact arrangement (HTHTH). Here W = number of microstates for a macrostate (not work!).
- Every microstate is assumed equally likely, so macrostates with more microstates are more likely and have higher entropy.
- Number of microstates for n heads in N coins = N! / (n!(N − n)!). Total for N coins = 2ᴺ.
| 5 coins |
5H |
4H 1T |
3H 2T |
2H 3T |
1H 4T |
5T |
Total |
| Microstates W |
1 |
5 |
10 |
10 |
5 |
1 |
32 |
- 100 coins: 1.27 × 10³⁰ microstates in all; 50–50 has 1.0 × 10²⁹, all heads has just 1. Tossing once a second, you'd wait about 2 × 10²² years to see 100 heads.
- Gas: the atoms spread out randomly because that macrostate has vastly more microstates. All the gas in one corner is so unlikely it's effectively impossible.
Ex. 15.9: 100 coins go from 60 H/40 T (W = 1.4 × 10²⁸) to 50 H/50 T (W = 1.0 × 10²⁹): ΔS = (1.38 × 10⁻²³)[ln(1.0 × 10²⁹) − ln(1.4 × 10²⁸)] = 2.7 × 10⁻²³ J/K.
Common mistakes (Chapter 15): sign of W (positive when the gas expands and does work; negative when compressed); sign of Q (positive into the system); forgetting ΔE_int = 0 for a full cycle or an isothermal ideal gas; using °C instead of kelvin in Eff_C, COP or ΔS = Q/T; mixing up efficiency (W/Q_h, always < 1) with **COP** (Q_h/W or Q_c/W, usually > 1); using COP_hp for a fridge (fridge uses Q_c); confusing W for work with W for number of microstates.
Practice 15.7 in the quiz → · Top ↑
Chapter 16: Oscillatory Motion and Waves
16.1 Hooke's Law: Stress and Strain Revisited
- Restoring force — a force that pushes an object back toward equilibrium (where the net force is zero). Pull a ruler left, it pushes right.
- Hooke's law — the restoring force is proportional to the displacement x. The minus sign means the force points opposite the displacement.
- Force constant k (N/m) — how stiff the system is. Bigger k = stiffer spring = bigger force for the same stretch.
- On a graph of |F| vs. x, a straight line means Hooke's law holds, and the slope is k.
- Newton's third law: the force you apply (F_app = kx) is equal and opposite to the restoring force.
- Elastic potential energy — the work you do deforming the system is stored as ½kx². It's the area of the triangle under the F vs. x graph.
- Damping (friction) removes energy so the motion eventually stops. Bouncy cars mean bad shock absorbers.
Ex. 16.1: a car settles 1.20 cm when an 80.0 kg person gets in: F = 80.0 × 9.80 = 784 N, k = 784 / 0.0120 = 6.53 × 10⁴ N/m.
Ex. 16.2: a toy gun spring (k = 50.0 N/m) compressed 0.150 m stores PE = ½(50.0)(0.150)² = 0.563 J, and launches a 2.00 g dart at v = √(2 × 0.563 / 0.00200) = 23.7 m/s.
Practice 16.1 in the quiz → · Top ↑
16.2 Period and Frequency in Oscillations
- Periodic motion — motion that repeats at regular time intervals (a guitar string, a mass on a spring).
- Period (T) — the time for one complete cycle, in seconds.
- Frequency (f) — the number of cycles per second, in hertz (Hz).
- They are opposites: long period = low frequency. Don't mix them up: T is "seconds per cycle," f is "cycles per second."
- If something does N cycles in time t: T = t/N and f = N/t.
| Problem |
Setup |
Answer |
| Ultrasound, T = 0.400 µs (Ex. 16.3a) |
f = 1 / 0.400 × 10⁻⁶ s |
2.50 × 10⁶ Hz |
| Middle C, f = 264 Hz (Ex. 16.3b) |
T = 1 / 264 |
3.79 × 10⁻³ s = 3.79 ms |
Practice 16.2 in the quiz → · Top ↑
16.3 Simple Harmonic Motion: A Special Periodic Motion
- Simple harmonic motion (SHM) — oscillation where the net force follows Hooke's law. The system is a simple harmonic oscillator.
- Amplitude (X) — the maximum displacement from equilibrium. With no damping, it goes the same distance on both sides.
- Period does NOT depend on amplitude. A guitar string plucked softly or hard plays the same note. That's why oscillators make good clocks.
- Only m and k set the period: heavier → slower (longer T); stiffer → faster (shorter T).
- The square root matters: 4× the mass only doubles the period.
- Graphing x vs. t gives a sine/cosine wave, which links SHM to waves.
- Speed is max at equilibrium and zero at the ends. Acceleration is max at the ends and zero at equilibrium, always pointing back toward the middle.
Ex. 16.4: a 900 kg car with k = 6.53 × 10⁴ N/m (bad shocks): f = (1/2π)√(6.53 × 10⁴ / 900) = 1.36 Hz, T = 1/f = 0.738 s.
Practice 16.3 in the quiz → · Top ↑
16.4 The Simple Pendulum
- Simple pendulum — a small heavy bob on a light string of length L.
- The restoring force is −mg sin θ. For angles under about 15°, sin θ ≈ θ, so the pendulum is a simple harmonic oscillator (with k = mg/L).
- Period depends only on L and g. It does NOT depend on the mass of the bob, and barely depends on amplitude (for small angles).
- Longer string → longer period. 4× the length → 2× the period.
- Weaker gravity (the Moon) → longer period. Measuring T and L is an accurate way to measure g.
- L is measured to the center of the bob.
Ex. 16.5: a 75.000 cm pendulum has a period of 1.7357 s: g = 4π²(0.75000) / (1.7357)² = 9.8281 m/s².
Practice 16.4 in the quiz → · Top ↑
16.5 Energy and the Simple Harmonic Oscillator
- With no damping, total energy stays constant and swaps back and forth between KE and PE_el.
- At the ends (x = ±X): all PE, v = 0. At equilibrium (x = 0): all KE, v = v_max.
- Total energy = ½kX², set by the amplitude.
- v_max goes up with amplitude, up with stiffness (√k), and down with mass (1/√m).
- For a pendulum: ω_max = √(g/L) θ_max (θ in radians).
Ex. 16.6: the 900 kg car (k = 6.53 × 10⁴ N/m) bounces with amplitude 0.100 m: v_max = √(6.53 × 10⁴ / 900) × 0.100 = 0.852 m/s.
Practice 16.5 in the quiz → · Top ↑
16.6 Uniform Circular Motion and Simple Harmonic Motion
- The shadow (projection) of an object moving in a circle at constant speed does simple harmonic motion along a line.
- The circle's radius is the amplitude X. The object's speed around the circle is v_max.
- Angular velocity ω (rad/s) = 2π/T = 2πf.
- One trip around the circle = one period. This gives the same T = 2π√(m/k) as before.
- Example: a ball on a turntable lit from the side makes a shadow that slides back and forth like a mass on a spring.
Practice 16.6 in the quiz → · Top ↑
16.7 Damped Harmonic Motion
- Damping — friction or other non-conservative forces remove mechanical energy (usually as heat). W_nc is negative.
- Light damping: amplitude slowly shrinks, but the period stays nearly the same.
| Type |
What happens |
Example |
| Underdamped |
Oscillates, crossing equilibrium many times while the amplitude shrinks |
Guitar string, playground swing |
| Critically damped |
Returns to equilibrium as fast as possible without oscillating (overshoots at most once) |
Good car shocks, bathroom scale needle |
| Overdamped |
Creeps slowly back to equilibrium without oscillating |
Door closer set too stiff |
Ex. 16.7: a 0.200 kg object on a spring (k = 50.0 N/m) with μ_k = 0.0800 feels friction f = 0.0800 × 0.200 × 9.80 = 0.157 N. Released 0.100 m from equilibrium, it travels a total of 1.59 m before stopping.
Practice 16.7 in the quiz → · Top ↑
16.8 Forced Oscillations and Resonance
- Natural frequency (f₀) — the frequency a system oscillates at on its own, with no driving or damping.
- Driving force — a periodic push from outside, at any frequency.
- Resonance — driving at the natural frequency. Energy transfers best, and the amplitude grows big.
- Less damping → taller and narrower resonance peak (piano strings, radio tuners). More damping → smaller amplitude but responds over a wider range of frequencies (car suspension).
- Examples: pushing a swing in time, singing a note to make a piano string ring or shatter a glass, radio tuning, MRI (~100 MHz), washboard gravel roads, and the 1940 Tacoma Narrows Bridge collapse.
Practice 16.8 in the quiz → · Top ↑
16.9 Waves
- Wave — a disturbance that travels from where it was created. The energy moves; the medium (water, string, air) mostly stays in place. A gull on the water just bobs up and down.
- Wavelength (λ) — the distance between neighboring identical parts (crest to crest), in meters.
- Wave velocity (v_w) — how fast the disturbance moves. One wavelength per period. v_w = fλ works for all waves.
- Amplitude is independent of wave speed.
- Transverse wave — disturbance perpendicular to travel (guitar strings, light, S-waves).
- Longitudinal wave — disturbance parallel to travel (sound in air and water, P-waves). Fluids can't support shear, so sound in them is longitudinal.
- Water waves are a mix of both. Earthquakes have both P (longitudinal) and S (transverse) waves.
Ex. 16.8: ocean wave crests 10.0 m apart; a gull bobs once every 5.00 s: v_w = λ/T = 10.0 / 5.00 = 2.00 m/s.
Practice 16.9 in the quiz → · Top ↑
16.10 Superposition and Interference
- Superposition — when waves meet, their disturbances add.
- Pure constructive interference — crest meets crest (in phase): amplitude doubles, same wavelength.
- Pure destructive interference — crest meets trough (out of phase): they cancel to zero.
- Standing wave — two identical waves moving in opposite directions; it looks like it vibrates in place (guitar strings, a glass of milk on a fridge).
- Node — a point that never moves (the fixed ends of a string). Antinode — the point of maximum amplitude.
- On a string fixed at both ends, the longest wavelength is λ₁ = 2L, giving the fundamental frequency f₁ = v_w/2L. The overtones (harmonics) are 2f₁, 3f₁, …
- Tighter string → faster v_w → higher pitch (that's how you tune a guitar).
- Beats — two waves with close frequencies go in and out of phase, so loudness throbs at the beat frequency f_B = |f₁ − f₂|. You hear the average frequency.
- Piano tuners adjust until the beats disappear (f_B = 0).
Ex. (beats): a 256 Hz tuning fork and a string give 2 beats per second, so the string is 254 Hz or 258 Hz.
Practice 16.10 in the quiz → · Top ↑
16.11 Energy in Waves: Intensity
- All waves carry energy. Energy ∝ amplitude squared: double the amplitude → 4× the energy and intensity.
- Intensity (I) — power per unit area, in W/m². Spreading a wave over more area lowers its intensity; focusing it raises it.
- Sunlight: about 1300 W/m² above the atmosphere, about 700 W/m² at the ground. A 90 dB sound is only 10⁻³ W/m².
- Two identical waves interfering constructively give 4× the intensity at that spot, but other spots get zero, so energy is still conserved.
| Problem |
Setup |
Answer |
| 700 W/m² on 0.500 m² for 4.00 h (Ex. 16.9a) |
E = 700 × 0.500 × (4.00 × 3600) |
5.04 × 10⁶ J |
| Same sunlight focused to 1/200 the area (Ex. 16.9b) |
I′ = 200 × 700 |
1.40 × 10⁵ W/m² |
| Two 1.00 W/m² waves, perfect constructive (Ex. 16.10) |
amplitude 2X → intensity (2)² = 4× |
4.00 W/m² |
Common mistakes (Chapter 16): mixing up period and frequency (T = 1/f; seconds vs. Hz); thinking a bigger amplitude changes the period (it doesn't, for springs or small-angle pendulums); thinking a heavier bob changes a pendulum's period (only L and g matter); forgetting the square root in T = 2π√(m/k) or 2π√(L/g); forgetting 2π; using cm instead of m for x or L; thinking intensity ∝ amplitude (it's ∝ amplitude²); thinking the water moves along with a wave (only the energy does); using λ = L instead of λ = 2L for a string's fundamental.
Practice 16.11 in the quiz → · Top ↑
Chapter 17: Physics of Hearing
17.1 Sound
- Sound — a disturbance of matter (a wave) that travels outward from its source. It exists whether or not anyone hears it.
- Hearing — the perception of sound, the way vision is the perception of light.
- In air and other fluids, sound is a longitudinal wave: the air moves back and forth along the direction the wave travels.
- A vibrating string or speaker makes compressions (slightly higher pressure) and rarefactions (slightly lower pressure). They move out at the speed of sound with the same frequency as the source.
- Fluids have almost no shear strength, so sound in them can't be transverse. (Solids can carry both kinds.)
- Sound gets weaker with distance because its energy spreads over a bigger area, and some is absorbed and turned into thermal energy.
- Pressure changes in ordinary sounds are tiny compared with atmospheric pressure.
Practice 17.1 in the quiz → · Top ↑
17.2 Speed of Sound, Frequency, and Wavelength
- Pitch — how we perceive frequency. High pitch = high frequency = short wavelength (piccolo). Low pitch = long wavelength (tuba).
- Wavelength (λ) — distance between neighboring compressions.
- The speed depends on the medium, not on the frequency. All notes from a marching band reach you together.
- Stiffer (less compressible) media carry sound faster. Gases are slowest, solids fastest. For similar stiffness, the lower-density material is faster.
- In air, speed grows with the square root of kelvin temperature: 331 m/s at 0°C, 343 m/s at 20°C.
- When sound moves into a new medium, frequency stays the same and the wavelength changes.
- Echoes: the sound goes there and back, so distance = v × t / 2. Bats use this.
- Earthquakes: P-waves (longitudinal, 4–7 km/s) beat S-waves (transverse, 2–5 km/s). The time gap gives the distance to the epicenter.
| Medium |
v (m/s) |
| Air (0°C) |
331 |
| Air (20°C) |
343 |
| Carbon dioxide (0°C) |
259 |
| Helium (0°C) |
965 |
| Hydrogen (0°C) |
1290 |
| Fresh water (20°C) |
1480 |
| Seawater / human tissue |
1540 |
| Vulcanized rubber |
54 |
| Lead |
1960 |
| Marble |
3810 |
| Aluminum |
5120 |
| Pyrex glass |
5640 |
| Steel |
5960 |
Ex. 17.1: at 30.0°C, v = 331√(303/273) = 348.7 m/s. Wavelengths: 348.7/20 = 17 m and 348.7/20,000 = 1.7 cm.
Practice 17.2 in the quiz → · Top ↑
17.3 Sound Intensity and Sound Level
- Intensity (I) — power per area, in W/m². It goes with the pressure amplitude squared (Δp is half the difference between the max and min pressure).
- Sound intensity level (β) — in decibels (dB). It compares I to the threshold of hearing I₀. No real units; "dB" just reminds you of the ×10.
- 0 dB is the threshold of hearing, not "no sound".
- Every 10 dB = 10× the intensity. +3 dB = 2× the intensity. +7 dB = 5×. 90 dB vs. 60 dB = 10³ = 1000× as intense.
- To go backward: I = I₀ × 10^(β/10).
- Over about 85 dB for 8 hours a day can damage hearing.
| β (dB) |
I (W/m²) |
Example |
| 0 |
10⁻¹² |
Threshold of hearing at 1000 Hz |
| 10 |
10⁻¹¹ |
Rustle of leaves |
| 20 |
10⁻¹⁰ |
Whisper at 1 m |
| 40 |
10⁻⁸ |
Average home |
| 60 |
10⁻⁶ |
Normal conversation |
| 70 |
10⁻⁵ |
Busy traffic |
| 80 |
10⁻⁴ |
Loud radio, classroom lecture |
| 90 |
10⁻³ |
Inside a heavy truck |
| 100 |
10⁻² |
Noisy factory, siren at 30 m |
| 120 |
1 |
Loud rock concert; threshold of pain |
| 140 |
10² |
Jet at 30 m; damage in seconds |
| 160 |
10⁴ |
Eardrums burst |
| I₂/I₁ |
β₂ − β₁ |
| 2 |
3 dB |
| 5 |
7 dB |
| 10 |
10 dB |
Ex. 17.2: Δp = 0.656 Pa in 0°C air: I = 0.656² / (2 × 1.29 × 331) = 5.04 × 10⁻⁴ W/m², so β = 10 log(5.04 × 10⁸) = 87 dB.
Ex. 17.3: twice the intensity: β₂ − β₁ = 10 log(2.00) = 3.01 dB.
Practice 17.3 in the quiz → · Top ↑
17.4 Doppler Effect and Sonic Booms
- Doppler effect — the observed frequency changes because the source or the observer moves. The Doppler shift is the size of that change.
- Toward each other → higher pitch. Apart → lower pitch. Faster motion, bigger shift.
- Sign rule: pick the sign that makes the answer go the right way. Moving source: − when approaching, + when leaving. Moving observer: + when approaching, − when leaving.
- The shifts are not symmetric: the rise on approach is bigger than the drop going away.
- If source and observer move together (the train's own engineer), there's no shift.
- Sonic boom — when the source goes faster than sound, the waves pile up into a cone-shaped shock wave of constructive interference. Planes make two booms (nose and tail). You hear it after the plane passes, and it lasts as long as the plane is supersonic.
- Bow wake — the same idea for water waves (the V behind a duck or boat). Cerenkov radiation is the light version.
- Uses: police radar, Doppler weather radar, blood-flow ultrasound, galaxy red shift.
Ex. 17.4: a 150 Hz horn on a train at 35.0 m/s (v = 340 m/s): approaching 150 × 340/305 = 167 Hz; leaving 150 × 340/375 = 136 Hz. The engineer hears 150 Hz.
Practice 17.4 in the quiz → · Top ↑
17.5 Sound Interference and Resonance: Standing Waves in Air Columns
- Interference proves something is a wave. Noise-canceling headphones make a sound with its highs and lows reversed (destructive interference), cutting noise 30 dB or more.
- Beats — two close frequencies together throb at f_B = |f₁ − f₂|.
- Resonance — only certain natural frequencies build standing waves in a tube; others cancel out.
- Node — no air motion (at a closed end). Antinode — biggest air motion (at an open end).
- Fundamental — the lowest resonant frequency (first harmonic). Overtones — the higher ones. Harmonics — all resonant frequencies, whole-number multiples of the fundamental.
| Tube |
Ends |
Fundamental λ |
Harmonics present |
Examples |
| Closed at one end |
Node + antinode |
4L |
Odd only: f, 3f, 5f… |
Clarinet, trumpet, your throat, a bottle |
| Open at both ends |
Antinode + antinode |
2L |
All: f, 2f, 3f… |
Flute, some organ pipes, oboe |
- For the same length, an open tube's fundamental is twice a closed tube's.
- Closed tube: the 1st overtone is the 3rd harmonic. Open tube: the 1st overtone is the 2nd harmonic.
- The mix of overtones is why a trumpet and a clarinet sound different on the same note.
- Resonant frequencies depend on the speed of sound, so on temperature. Musicians warm up their instruments.
Ex. 17.5: a closed tube with a 128 Hz fundamental at 22.0°C: v = 331√(295/273) = 344 m/s, L = 344 / (4 × 128) = 0.672 m. Its 4th overtone is n = 9: 9 × 128 = 1.15 kHz.
Practice 17.5 in the quiz → · Top ↑
17.6 Hearing
- Human hearing: 20 to 20,000 Hz. Below 20 Hz is infrasound (elephants); above 20,000 Hz is ultrasound (dogs to 30 kHz, bats and dolphins to 100 kHz).
- We can tell apart frequencies that differ by about 0.3%. Perfect pitch is naming a note by ear.
- Loudness — perception of intensity, but it also depends on frequency. We notice about 1 dB; 3 dB is easy to notice.
- The ear is most sensitive from 2000 to 5000 Hz (partly from resonance in the ear canal).
- Phon — the unit of loudness. Equal-loudness curves show sounds that seem equally loud. At 1000 Hz, phons = decibels.
- Timbre (tone quality) — the mix of frequencies and their intensities. It's why a violin and a piano sound different on middle C.
| Perception |
Physical quantity |
| Pitch |
Frequency |
| Loudness |
Intensity and frequency |
| Timbre / tone |
Number and relative intensity of the frequencies |
- Ear parts: outer ear (ear canal) → middle ear (eardrum and the tiny bone lever) → inner ear (cochlea). The middle ear's lever makes pressure in the cochlea about 40× larger than at the eardrum.
- In the cochlea, hair cells (cilia) turn vibrations into nerve signals. High frequencies trigger the near end; low frequencies the far end.
- Noise damage shows as a dip near 4000 Hz. Presbycusis is age-related loss of high frequencies.
- Hearing tests (audiograms) run from 250 to 8000 Hz. Cochlear implants stimulate the nerves directly.
Ex. 17.6: from the equal-loudness graph, a 100 Hz sound at 80 dB has a loudness of 75 phons; a 4000 Hz sound at 70 phons is only 67 dB.
Practice 17.6 in the quiz → · Top ↑
17.7 Ultrasound
- Ultrasound — above 20 kHz. Made and detected by a piezoelectric transducer (it works as both speaker and microphone).
- Acoustic impedance (Z) — density × speed of sound, in kg/(m²·s).
- Intensity reflection coefficient (a) — the fraction of intensity reflected at a boundary. Big difference in Z → big reflection. Equal Z → no reflection. That's why gel is used: it removes the air gap (air's Z is tiny).
- Imaging: echo time tells depth. You can't see details smaller than about one wavelength, and you can only scan to about 500λ deep. Higher frequency = more detail but less depth.
- Doppler ultrasound measures blood speed. There's a double shift (blood receives as a moving observer, then re-sends as a moving source). The beat frequency between sent and returning waves gives the speed.
- Therapy: 10³–10⁵ W/m² can shatter gallstones; diathermy (deep heating) at 10³–10⁴ W/m². Diagnostic scans use about 10⁻² W/m², which is safe.
- Cavitation — vapor bubbles form and collapse. Used in ultrasonic cleaners (about 40 kHz).
- Other uses: sonar (30–100 kHz), ranging devices, finding cracks in aircraft wings.
| Medium |
ρ (kg/m³) |
v (m/s) |
Z (kg/(m²·s)) |
| Air |
1.3 |
330 |
429 |
| Water |
1000 |
1500 |
1.5 × 10⁶ |
| Blood |
1060 |
1570 |
1.66 × 10⁶ |
| Fat |
925 |
1450 |
1.34 × 10⁶ |
| Muscle |
1075 |
1590 |
1.70 × 10⁶ |
| Bone |
1400–1900 |
4080 |
5.7–7.8 × 10⁶ |
| Problem |
Setup |
Answer |
| Fat to muscle (Ex. 17.7) |
Z_fat = 925 × 1450; a = (1.34 − 1.70)² / (1.34 + 1.70)² |
1.34 × 10⁶; a = 0.014 (1.4% reflected) |
| 7 MHz scan in tissue |
λ = 1540 / 7 × 10⁶; depth = 500λ |
0.22 mm detail; 0.11 m deep |
| 2.50 MHz on blood at 20.0 cm/s toward probe (Ex. 17.8) |
Blood receives 2,500,325 Hz; echo returns 2,500,649 Hz |
Beat frequency 649 Hz |
Common mistakes (Chapter 17): treating decibels as linear (doubling intensity adds 3 dB, it doesn't double the dB); using ln instead of log₁₀; using °C instead of kelvin in v = 331√(T/273); thinking higher frequency travels faster (speed depends on the medium only); using 2L for a closed tube (it's 4L) or allowing even harmonics in a closed tube; flipping the Doppler sign (approaching must give a higher frequency); forgetting that an echo travels there and back.
Practice 17.7 in the quiz → · Top ↑