Unit 3: Work, Energy & Power · First Principles Physics
AP Physics 1 · Unit 03 of 08

Work, Energy & Power.

Newton's laws can solve any mechanics problem in principle. But for the messy ones — collisions, complicated paths, springs — energy methods are dramatically faster. This unit gives you a second toolbox, and the right one to reach for most of the time.

Exam Weight · 18–23% Class Periods · ~22–27 CED Topics · 3.1–3.5 Difficulty · Moderate
01
The lesson

A second way to solve every mechanics problem — usually faster

A roller coaster crests the top of its first hill at 5 m/s. The track twists down a 40-meter drop, around a banked curve, up a smaller hill, then through a corkscrew. How fast is it going at the bottom? You could, in principle, set up Newton's second law along every centimeter of track, integrate forces over the curving path, and emerge with the answer. You'd be at it for hours. Or you could use this unit's tools — write down conservation of energy, plug in two numbers, and have the answer in fifteen seconds: 28 m/s.

That's the power of energy methods. They don't tell you anything about when things happen or how they got there — Newton's laws still own that territory. But for questions of the form "how fast?" or "how high?" or "how far?", energy is the way. This unit is also the spiritual heart of physics: the principle of energy conservation, which Einstein called "the most fundamental law of nature," is the closest thing physics has to a sacred text.

The key insight: instead of tracking forces moment by moment, you track a single number — the total energy — that doesn't change as a system moves through complicated configurations. This single conservation law turns hard problems into easy ones.

Kinetic energy

An object's kinetic energy is the energy of its motion:

$$K = \tfrac{1}{2}mv^2$$

Units: joules (J = kg·m²/s²). Some things to notice. Kinetic energy is a scalar — it has a magnitude but no direction. Doubling speed quadruples KE. A truck going 30 mph has 25 times the KE of a bicycle at the same speed (assuming the truck is 25× heavier). And, importantly, KE depends on the observer's reference frame — a passenger sitting still in a car has zero KE relative to the driver but enormous KE relative to a person standing on the sidewalk.

Work — how energy gets transferred

Energy doesn't materialize from nowhere; it gets transferred. The mechanism for transferring energy via a force is called work:

$$W = F\,d\cos\theta$$

Here \(F\) is the magnitude of the force, \(d\) is the magnitude of the displacement, and \(\theta\) is the angle between them. Work is also a scalar, measured in joules. Three special cases worth memorizing:

  • Force parallel to motion (\(\theta = 0°\)): \(W = +Fd\). Force does positive work, transfers energy into the object.
  • Force opposite to motion (\(\theta = 180°\)): \(W = -Fd\). Negative work, energy comes out of the object.
  • Force perpendicular to motion (\(\theta = 90°\)): \(W = 0\). No energy transfer.

That last one is non-obvious and crucial. The normal force on a block sliding across a horizontal floor does no work on the block, even though the normal force is large. Why? It's perpendicular to the block's motion. Same for the centripetal force on an object moving in a circle at constant speed — perpendicular to motion, no work done, kinetic energy unchanged. The string tension on a rock swung in a circle isn't speeding the rock up; it's just changing its direction.

The work-energy theorem

The connection between work and kinetic energy is one of the cleanest results in physics. The total (net) work done on an object equals the change in its kinetic energy:

$$W_{\text{net}} = \Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$$

This gives you a powerful alternative to Newton's second law. If you know all the forces and how far each acts, you can compute the total work, and that gives you the change in KE — which gives you the final speed. No time-keeping required. This is why energy methods shine in problems where forces vary, paths curve, or you simply don't care about when things happen.

Worked example 1 · work-energy theorem

Stopping a sliding block

A 4 kg block slides at 6 m/s onto a rough surface where \(\mu_k = 0.3\). How far does it slide before stopping? Use \(g = 10\) m/s².

Identify the forces and work

Friction \(f_k = \mu_k mg = 0.3(40) = 12\) N opposes motion. Over distance \(d\), friction does work \(W = -f_k d = -12d\).

Apply work-energy theorem

Block stops, so \(v_f = 0\). Initial KE: \(\tfrac{1}{2}(4)(36) = 72\) J. Final KE: 0.

\(W_{\text{net}} = \Delta K \Rightarrow -12d = 0 - 72 \Rightarrow d = 6\) m.

\(d = 6\) m

Potential energy

Some forces have a special property: the work they do on an object only depends on the object's start and end positions, not on the path it took. These are called conservative forces. Gravity is conservative — lift a book up 1 meter and the work done by gravity is \(-mg(1\text{ m})\), regardless of whether you took it straight up or along a winding path. Spring forces are conservative. Friction is not — it always opposes motion, so the work it does depends on the path length.

For each conservative force, we can define a potential energy \(U\) such that the force is "stored" in the object's position. Work done by gravity equals minus the change in gravitational PE; same for springs. Two specific formulas you must know:

$$U_g = mgh$$ near Earth's surface
$$U_s = \tfrac{1}{2}kx^2$$ spring stretched/compressed by x

Notice the spring PE is \(\tfrac{1}{2}kx^2\), not \(kx^2\) — the half comes from integrating the linear spring force over the displacement. This is one of the most commonly forgotten factors in Unit 3.

For systems where objects are very far apart (say, a rocket far from Earth), the constant-\(g\) approximation breaks down and we use the full Newtonian formula:

$$U_g = -\dfrac{Gm_1 m_2}{r}$$

The minus sign and the choice of \(U = 0\) at \(r = \infty\) are conventions — they make \(U\) negative for bound systems, which turns out to be useful for thinking about escape velocity and orbits.

Conservation of mechanical energy

The total mechanical energy of a system is the sum of its kinetic and potential energies:

$$E = K + U$$

If the only forces doing work on a system are conservative (gravity, springs), then the mechanical energy is constant — energy can transform between kinetic and potential, but the total stays the same:

$$K_i + U_i = K_f + U_f$$

This is conservation of mechanical energy, and it's the workhorse equation of Unit 3. Drop a ball from height \(h\): initial KE is 0, initial PE is \(mgh\), final PE (just before hitting the ground) is 0, so final KE must be \(mgh\), giving \(v = \sqrt{2gh}\). Pluck a vertical spring with mass attached, find max compression: spring PE \(\tfrac{1}{2}kx_{\max}^2\) equals the gravitational PE the mass loses dropping that distance, plus initial KE. Roller coaster crests a hill at 5 m/s, drops 40 m: \(\tfrac{1}{2}v_f^2 = \tfrac{1}{2}v_i^2 + g\Delta h\), so \(v_f = \sqrt{25 + 800} \approx 28\) m/s. Done in fifteen seconds.

When mechanical energy is NOT conserved

If a non-conservative force (most commonly friction) does work on the system, mechanical energy decreases. Friction converts mechanical energy to thermal energy (heat) — the block and surface get slightly warmer. The energy doesn't disappear; it just leaves the "mechanical" category. The accounting equation becomes \(K_i + U_i = K_f + U_f + |W_{\text{friction}}|\), where \(|W_{\text{friction}}| = f_k d\) is the energy lost to friction over a path of length \(d\).

Worked example 2 · energy conservation with spring

A block compressing a spring on a ramp

A 2 kg block is released from rest at the top of a frictionless 30° ramp, 5 m up the slope from the bottom of an uncompressed spring with \(k = 400\) N/m. How much does the spring compress at the block's lowest point?

Choose a reference

Set \(U_g = 0\) at the position where the block first touches the spring. The block is initially at height \(h_i = 5\sin 30° = 2.5\) m above this reference; at the moment of maximum compression, it's a distance \(x\) further down the slope, at height \(-x\sin 30° = -0.5x\) below.

Apply energy conservation

Initial: KE = 0, \(U_g = mgh_i = 2(10)(2.5) = 50\) J, \(U_s = 0\). Final (max compression): KE = 0, \(U_g = mg(-x\sin\theta) = -10x\), \(U_s = \tfrac{1}{2}(400)x^2 = 200 x^2\).

Conservation: \(50 = -10x + 200 x^2 \Rightarrow 200 x^2 - 10x - 50 = 0 \Rightarrow 20 x^2 - x - 5 = 0\).

Quadratic formula: \(x = (1 \pm \sqrt{1 + 400})/40 \approx (1 + 20.02)/40 \approx 0.526\) m.

Spring compresses by approximately 0.53 m.

Energy bar charts

The 2025 AP Physics 1 exam included a free-response question that asked students to draw energy bar charts at various points during a block-spring-ramp problem. Bar charts are simple: for each instant, draw a vertical bar for KE, one for gravitational PE, and one for spring PE, with the bar heights proportional to the magnitudes. The key observation: the total height of all bars must be the same at every instant (assuming no friction), because mechanical energy is conserved. As the block slides, energy moves between bars but the total stays constant.

Bar charts are useful both as a problem-solving tool (force you to think about every form of energy) and as a way to communicate your reasoning on the AP exam. Practice them.

Power

Energy is the amount transferred. Power is the rate:

$$P = \dfrac{W}{t} = Fv\cos\theta$$

Units: watts (W = J/s). A 100-watt lightbulb consumes 100 J of energy every second. Power is what your electric bill measures (in kilowatt-hours, which is energy = power × time, not power itself — common source of confusion). The same amount of work done in less time requires more power. The same engine power can move a heavy object slowly or a light object quickly.

The form \(P = Fv\) is useful when a constant force acts on a moving object — the instantaneous power is force times velocity. A car moving at 30 m/s with engine providing 1500 N of forward thrust delivers 45,000 W = 45 kW = about 60 horsepower.

Choosing systems wisely

One subtle but powerful idea in Unit 3 is the importance of system definition. Whether energy is "conserved" depends on what you call your system. If you analyze just a block sliding down a frictionless ramp, gravity does external work on the block — KE changes. But if you analyze the block + Earth as your system, gravity is now an internal force, and the energy just trades between KE (of the block) and gravitational PE (of the system). Mechanical energy is conserved.

The same is true for springs: include the spring in your system and you can use \(\tfrac{1}{2}kx^2\) as PE. Exclude it and the spring force does external work on the object. AP problems often hinge on this distinction. When in doubt, include the source of any conservative force in your system, and use PE accounting.

Where this unit goes next

Unit 4 introduces momentum — another conserved quantity, but for collisions and interactions specifically. Together, energy and momentum form the most powerful problem-solving toolkit in classical mechanics. Then Unit 5 will introduce the rotational versions (rotational KE, work by torque), so what you learn here is half of two later units. Master conservation of energy thoroughly; it pays dividends for the rest of the course.

Conceptual summary

Energy methods provide a faster alternative to Newton's laws for problems involving speeds, heights, or distances. Kinetic energy is the energy of motion (\(\tfrac{1}{2}mv^2\)); potential energy is the stored energy of position (gravitational \(mgh\) or spring \(\tfrac{1}{2}kx^2\)). Work is energy transfer via force: \(W = Fd\cos\theta\). The work-energy theorem says net work equals change in KE. When only conservative forces act, mechanical energy (\(K + U\)) is conserved. Friction transfers mechanical energy to thermal — so we get \(\Delta E = -|W_{\text{friction}}|\). Power is the rate of energy transfer, \(P = W/t = Fv\). Choosing your system to include the source of conservative forces lets you use PE bookkeeping and the conservation principle.

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02
Reference sheet

Every equation in Unit 3

★ Translational kinetic energy

$$K = \tfrac{1}{2}mv^2$$
\(m\)
mass (kg)
\(v\)
speed (m/s)

Use when finding the energy of a moving object. Scalar — no direction.

★ Work by a force

$$W = Fd\cos\theta$$
\(F\)
force magnitude (N)
\(d\)
displacement magnitude (m)
\(\theta\)
angle between force and displacement

Use when a constant force acts over a displacement. Sign matters.

★ Work-energy theorem

$$W_{\text{net}} = \Delta K$$
\(W_{\text{net}}\)
sum of work done by all forces
\(\Delta K\)
change in KE: \(K_f - K_i\)

Use when finding speed change due to forces, especially when path is complex.

★ Gravitational PE (near Earth)

$$U_g = mgh$$
\(h\)
height above chosen reference

Use when object is near Earth's surface. Reference height is yours to choose.

★ Universal gravitational PE

$$U_g = -\dfrac{Gm_1 m_2}{r}$$
\(r\)
center-to-center separation
Sign
negative for bound systems

Use when objects are far from Earth's surface (orbits, escape velocity).

★ Spring PE

$$U_s = \tfrac{1}{2}kx^2$$
\(k\)
spring constant (N/m)
\(x\)
displacement from equilibrium

Use when a spring is stretched/compressed. Don't forget the ½.

★ Conservation of mechanical energy

$$K_i + U_i = K_f + U_f$$
Condition
only conservative forces do work

Use when friction is absent and you need final speed/height.

Conservation with friction

$$K_i + U_i = K_f + U_f + f_k d$$
\(f_k d\)
energy dissipated by friction over path of length \(d\)

Use when friction is present. The dissipated energy becomes thermal.

★ Power (average)

$$P = \dfrac{W}{t}$$
\(t\)
time elapsed (s)
Units
watts (W)

Use when finding rate of energy transfer over a time interval.

★ Power (instantaneous)

$$P = Fv\cos\theta$$
\(\theta\)
angle between force and velocity

Use when a constant force acts on a moving object. Often \(\theta = 0\), so \(P = Fv\).

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04
Worksheet A · Foundation

Ten problems building from basic to moderate

Use \(g = 10\) m/s² throughout. Reach for conservation of energy unless told otherwise.
Foundation

A 3 kg object moves at 4 m/s. Find its kinetic energy.

Foundation

A 50 N force is applied to a box. The box moves 3 m in the direction of the force. Find the work done.

Foundation

A 2 kg book is lifted 1.5 m off the floor. Find its change in gravitational potential energy.

Moderate

A 1.5 kg ball is dropped from rest at a height of 3 m. Using conservation of energy, find its speed just before hitting the ground.

Moderate

A spring with \(k = 200\) N/m is compressed 0.1 m. Find the spring's potential energy.

Moderate

A 100 N force is applied to a 10 kg block at a 30° angle above the horizontal. The block moves 4 m horizontally. Find the work done by the applied force.

Moderate

A 5 kg block slides at 8 m/s onto a horizontal surface where \(\mu_k = 0.4\). Using the work-energy theorem, find how far it slides before stopping.

Moderate

A roller coaster car (mass 500 kg) starts from rest at the top of a 30 m frictionless hill. Find its speed at the bottom.

Moderate

A 0.5 kg ball is launched horizontally from a spring (\(k = 800\) N/m) compressed 0.2 m on a frictionless table. Find the ball's speed when it leaves the spring.

Moderate

A motor lifts a 200 kg crate 5 m vertically at constant speed in 4 seconds. Find (a) the work done by the motor, (b) the average power output of the motor.

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05
Worksheet B · AP exam style

Ten problems in AP exam format

Part I — Multiple Choice

AP MC

A 2 kg block slides at 6 m/s. A second 2 kg block slides at 3 m/s. The kinetic energy ratio \(K_1/K_2\) is:

  1. 1
  2. 2
  3. 4
  4. 8
AP MC

A box slides across a horizontal floor at constant speed. Which of the following statements is correct?

  1. Net work on the box is positive.
  2. Net work on the box is negative.
  3. Net work on the box is zero.
  4. The applied force does no work.
AP MC

Block A (mass \(m\)) is dropped from height \(h\). Block B (mass \(2m\)) is dropped from height \(h/2\). Both are in free fall. The ratio of the speed of block A just before hitting the ground to the speed of block B is:

  1. 1 to 1
  2. \(\sqrt{2}\) to 1
  3. 2 to 1
  4. 4 to 1
AP MC

A spring is compressed 0.1 m and stores 5 J of potential energy. If instead it is compressed 0.3 m, the stored potential energy is:

  1. 15 J
  2. 25 J
  3. 45 J
  4. 75 J
AP MC

A pendulum swings back and forth. At which point is its kinetic energy at maximum?

  1. At the highest point of swing.
  2. At the lowest point of swing.
  3. Halfway between highest and lowest.
  4. The KE is constant throughout the swing.

Part II — Short Free Response

AP SFR

(~7 min) A 1 kg ball is dropped from rest at a height of 5 m above the floor. Air resistance is negligible.

  1. Derive a symbolic expression for the speed of the ball just before it hits the floor, starting from conservation of energy.
  2. Calculate the speed.
  3. Sketch a qualitative graph of the ball's KE and gravitational PE vs height, with both energies on the same axis. Label both curves.
AP SFR

(~7 min) A 4 kg block slides on a horizontal surface with coefficient of kinetic friction 0.25. Initially the block has speed 6 m/s, and a horizontal applied force of 30 N is exerted in the direction of motion.

  1. Calculate the work done by the applied force as the block moves 3 m.
  2. Calculate the work done by friction over the same 3 m.
  3. Use the work-energy theorem to find the block's speed after moving 3 m.
AP SFR

(~7 min) A 0.5 kg block on a frictionless horizontal surface is attached to a spring (\(k = 200\) N/m). The block is pushed 0.4 m from equilibrium and released.

  1. Calculate the maximum kinetic energy of the block.
  2. Calculate the maximum speed of the block.
  3. If the surface had friction with \(\mu_k = 0.2\), would the block return to its starting position with zero velocity? Justify your answer.

Part III — Extended Free Response

AP EFR

(~18 min — mirrors 2025 AP Physics 1 FRQ Q2) A block of mass \(M\) is released from rest at position \(x = 0\) at the top of a frictionless ramp making angle \(\theta\) with the horizontal. The block slides down and at \(x = 8D\) makes contact with an uncompressed spring with spring constant \(k\). The block compresses the spring and momentarily stops at \(x = 12D\).

  1. At \(x = 10D\) (the block is 4D below its starting position along the slope, with spring compressed \(2D\)), draw an energy bar chart showing relative magnitudes of kinetic energy \(K\), gravitational potential energy \(U_g\), and spring potential energy \(U_s\). Define \(U_g = 0\) at \(x = 12D\).
  2. Starting from conservation of energy, derive an expression for the spring constant \(k\) in terms of \(M\), \(\theta\), \(D\), and physical constants.
  3. On axes provided (Energy vs Position from \(8D\) to \(12D\)), sketch curves for total mechanical energy \(E\) and gravitational PE \(U_g\) of the block-spring-Earth system, given that the spring PE \(U_s\) has been pre-drawn.
  4. If the ramp now has friction (coefficient \(\mu_k\)), the block compresses the spring less than before. Derive a symbolic expression for the new compression distance.
AP EFR

(~18 min — power and energy) A 1500 kg car accelerates from rest along a horizontal road with a constant engine force of 4500 N. Friction and air drag combined produce an effective resistive force of 1500 N (independent of speed for this problem).

  1. Find the car's acceleration.
  2. Use the work-energy theorem to find the car's speed after it has traveled 100 m.
  3. Find the work done by the engine over those 100 m.
  4. Find the power delivered by the engine at the moment the car has reached the speed found in part (b).
  5. If the engine has a maximum power output of 60 kW, find the maximum speed the car can reach. Justify your answer.
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06
Worksheet C · Challenge

Five problems at olympiad level

Challenge

Block on a circular hill. A small block of mass \(m\) starts at rest at the top of a frictionless hemispherical hill of radius \(R\). It slides down the surface. Find the angle (measured from the top, vertical) at which the block leaves the surface. (Hint: the block leaves the surface when the normal force becomes zero.)

Challenge

Two-spring system. A 2 kg block on a frictionless horizontal surface is connected to two springs (each with \(k = 100\) N/m) attached to walls on opposite sides. The block is initially at rest with both springs at natural length. The block is then displaced 0.1 m to the right and released. (a) Find the maximum speed of the block. (b) The block oscillates back and forth — find the amplitude of oscillation. (c) If a third spring with \(k = 200\) N/m is added, attached to the block in series with the right-hand spring, find the new effective spring constant felt by the block. (Hint: springs in series.)

Challenge

Loop with friction patch. A small block of mass \(m\) starts from rest at height \(h\) on a frictionless track that ends in a vertical circular loop of radius \(R\). On the bottom horizontal section of the track, there's a patch of length \(L\) with kinetic friction coefficient \(\mu_k\). What minimum height \(h\) is required for the block to maintain contact at the top of the loop?

Challenge

Bungee jumper. A 60 kg bungee jumper falls from a high platform attached to a bungee cord with natural length 20 m and effective spring constant 50 N/m (when stretched). (a) Find the maximum stretch of the cord. (b) Find the jumper's maximum speed during the fall. (c) Compare your answer for max speed to the speed they would have hit the ground at without the cord, having fallen the same distance the cord eventually stretches. Briefly explain.

Challenge

Escape velocity. Starting from energy conservation and the universal gravitational PE \(U = -GMm/r\), derive the escape velocity from the surface of a planet of mass \(M\) and radius \(R\). Then calculate the escape velocity from Earth (\(M = 5.97 \times 10^{24}\) kg, \(R = 6.37 \times 10^6\) m, \(G = 6.67 \times 10^{-11}\) N·m²/kg²). Comment briefly on whether the answer matches the rule-of-thumb value of 11.2 km/s.

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07
Complete worked solutions

Every problem, every step

Worksheet A — Foundation

· KE calculation

\(K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(3)(16) = 24\) J.

\(K = 24\) J
· simple work

Force parallel to displacement (\(\theta = 0°\)): \(W = Fd = (50)(3) = 150\) J.

\(W = 150\) J
· gravitational PE change

\(\Delta U_g = mg\Delta h = (2)(10)(1.5) = 30\) J.

\(\Delta U_g = 30\) J
· dropped ball — speed at ground

Energy conservation: \(mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{60} \approx 7.75\) m/s.

\(v \approx 7.75\) m/s
Mass cancels — every dropped object reaches the same speed after falling the same height (in vacuum). This is the same result you'd get from kinematics, but reached in one line via energy.
· spring PE

\(U_s = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.01) = 1.0\) J.

\(U_s = 1.0\) J
Don't forget the ½. Writing \(kx^2 = 2.0\) J is the most common error in spring problems.
· work at an angle

\(W = Fd\cos\theta = (100)(4)\cos 30° = 400(0.866) \approx 346\) J.

\(W \approx 346\) J
· friction stopping distance

Friction does negative work: \(W_f = -\mu_k mg \cdot d = -0.4(50)d = -20d\). Block stops, so \(\Delta K = 0 - \tfrac{1}{2}(5)(64) = -160\) J. From WET: \(-20d = -160 \Rightarrow d = 8\) m.

\(d = 8\) m
· roller coaster down a hill

Energy conservation (frictionless): \(mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{600} \approx 24.5\) m/s.

\(v \approx 24.5\) m/s
Notice how mass and the shape of the hill don't appear — energy methods don't care whether the hill is straight, curved, twisted, or has a corkscrew. As long as it's frictionless, only the height drop matters.
· spring-launched ball

Spring PE → KE: \(\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2 \Rightarrow v = x\sqrt{k/m} = 0.2\sqrt{800/0.5} = 0.2\sqrt{1600} = 0.2(40) = 8\) m/s.

\(v = 8\) m/s
· motor lifting crate

(a) Constant speed: motor does work equal to gain in PE: \(W = mgh = (200)(10)(5) = 10{,}000\) J = 10 kJ. (b) Power: \(P = W/t = 10000/4 = 2500\) W = 2.5 kW.

\(W = 10\) kJ; \(P = 2.5\) kW.

Worksheet B — AP Exam Style

· KE ratio

Same mass; \(K \propto v^2\). \(v_1/v_2 = 6/3 = 2\), so \(K_1/K_2 = 4\).

(C) 4
· constant speed across floor

Constant speed → \(\Delta K = 0\) → \(W_{\text{net}} = 0\) by the work-energy theorem.

(C) Net work is zero.
The applied force does positive work, friction does equal-magnitude negative work, and the two cancel. Net work zero, but individual works are not zero.
· two free-fall blocks

Block A: \(v_A = \sqrt{2gh}\). Block B: \(v_B = \sqrt{2g(h/2)} = \sqrt{gh}\). Ratio: \(v_A/v_B = \sqrt{2}\).

(B) \(\sqrt{2}\) to 1
· spring PE scaling

\(U_s \propto x^2\). Compression triples (0.1 → 0.3), so PE multiplies by 9: 5 J → 45 J.

(C) 45 J
· pendulum max KE

Energy conservation: when PE is at minimum (lowest point), KE is at maximum.

(B) At the lowest point.
· dropped ball with graph

(a) Conservation: \(mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}\). (b) \(v = \sqrt{2(10)(5)} = 10\) m/s. (c) KE vs height: starts at 0 at \(h = 5\), increases linearly to \(mgh = 50\) J at \(h = 0\). PE vs height: starts at 50 J at \(h = 5\), decreases linearly to 0 at \(h = 0\). The two curves are mirror images, summing to a constant 50 J.

\(v = 10\) m/s; KE and PE are linear lines crossing at \(h = 2.5\), \(E = 25\) J each.
· block with applied force + friction

(a) \(W_{\text{app}} = (30)(3) = 90\) J. (b) \(f_k = \mu_k mg = 0.25(40) = 10\) N; \(W_f = -10(3) = -30\) J. (c) \(W_{\text{net}} = 90 - 30 = 60\) J. WET: \(60 = \tfrac{1}{2}(4)v_f^2 - \tfrac{1}{2}(4)(36) = 2v_f^2 - 72\). Solve: \(v_f^2 = 66 \Rightarrow v_f \approx 8.12\) m/s.

\(W_{\text{app}} = 90\) J; \(W_f = -30\) J; \(v_f \approx 8.12\) m/s.
· spring-block system

(a) Max KE = initial spring PE = \(\tfrac{1}{2}(200)(0.16) = 16\) J. (b) \(\tfrac{1}{2}mv^2 = 16 \Rightarrow v = \sqrt{64} = 8\) m/s. (c) No. With friction, energy dissipates each pass: \(f_k = \mu_k mg = 0.2(5) = 1\) N. After traveling 0.4 m back to start, energy lost is \(f_k \cdot 0.8 = 0.8\) J (full round trip). Block reaches start with KE \(\approx 16 - 0.8 = 15.2\) J still, so it overshoots and continues — does NOT come to rest at start.

Max KE = 16 J; max speed 8 m/s; with friction, block does not return to starting position with zero velocity.
· 2025-style block-spring-ramp EFR

(a) At \(x = 10D\): block has dropped \((12D - 10D)\sin\theta = 2D\sin\theta\) since the moment of contact, but is still 4D below starting point along slope. With reference at \(x = 12D\): \(U_g\) at \(x = 10D\) corresponds to height \(2D\sin\theta\) above reference. Spring compressed \(10D - 8D = 2D\), so \(U_s = \tfrac{1}{2}k(2D)^2 = 2kD^2\). Bar chart: KE bar (some height), \(U_g\) bar (some height for \(2D\sin\theta\) above reference), \(U_s\) bar (height for \(2kD^2\)). Total height equals total mechanical energy = \(MgD\sin\theta \cdot 12\) (from start) since \(U_g = 0\) reference is at end position.

(b) Energy conservation from start (\(x = 0\), at rest) to maximum compression (\(x = 12D\), at rest):

\(Mg(12D)\sin\theta = \tfrac{1}{2}k(4D)^2\) (compression is \(12D - 8D = 4D\))

\(12MgD\sin\theta = 8kD^2 \Rightarrow k = \dfrac{3Mg\sin\theta}{2D}\).

(c) Total energy \(E\) is constant — horizontal line. \(U_g\) decreases linearly with position (line from positive value at \(x = 8D\) to 0 at \(x = 12D\) for the chosen reference). The provided spring PE curve and these two together must satisfy \(K + U_g + U_s = E\).

(d) With friction (kinetic friction coefficient \(\mu_k\)), the block compresses the spring by some new amount \(d\). Energy balance: \(Mg(8D + d)\sin\theta = \tfrac{1}{2}kd^2 + \mu_k Mg\cos\theta(8D + d)\). Substitute \(k\) from part (b) and solve quadratic for \(d\).

\(k = 3Mg\sin\theta/(2D)\); friction case requires solving quadratic.
· car power and energy

(a) Net force: \(4500 - 1500 = 3000\) N. \(a = 3000/1500 = 2\) m/s².

(b) WET: \(W_{\text{net}} = 3000(100) = 300{,}000\) J = \(\tfrac{1}{2}(1500)v^2 \Rightarrow v^2 = 400 \Rightarrow v = 20\) m/s.

(c) Engine work: \(W_{\text{eng}} = (4500)(100) = 450{,}000\) J = 450 kJ.

(d) Power at \(v = 20\) m/s: \(P = Fv = (4500)(20) = 90{,}000\) W = 90 kW.

(e) If max engine power is 60 kW, max speed at full force is \(P/F = 60000/4500 \approx 13.3\) m/s. But this assumes the resistive force stays the same; at max speed the car must have zero net force, so engine force equals resistive force (1500 N). Max speed: \(P_{\max} = F_{\text{resist}} \cdot v_{\max} \Rightarrow v_{\max} = 60000/1500 = 40\) m/s.

\(a = 2\) m/s²; \(v = 20\) m/s; \(W_{\text{eng}} = 450\) kJ; \(P_{\text{at 20 m/s}} = 90\) kW; \(v_{\max} = 40\) m/s.
At top speed, the car isn't accelerating — engine power is just keeping pace with drag. \(P_{\max} = F_{\text{drag}} v_{\max}\), so \(v_{\max} = P_{\max}/F_{\text{drag}}\). This is why a car's top speed is limited by power, not raw engine force.

Worksheet C — Challenge

· block on hemispherical hill

Let \(\phi\) = angle from top. Energy conservation: \(mgR(1 - \cos\phi) = \tfrac{1}{2}mv^2 \Rightarrow v^2 = 2gR(1 - \cos\phi)\).

Centripetal at angle \(\phi\): inward (toward center) component of net force = \(mv^2/R\). Forces: gravity has inward component \(mg\cos\phi\); normal points outward (\(N\)). So \(mg\cos\phi - N = mv^2/R\).

Block leaves when \(N = 0\): \(mg\cos\phi = mv^2/R = 2mg(1 - \cos\phi) \Rightarrow \cos\phi = 2 - 2\cos\phi \Rightarrow 3\cos\phi = 2 \Rightarrow \cos\phi = 2/3\), so \(\phi \approx 48.2°\).

Block leaves at \(\phi = \arccos(2/3) \approx 48.2°\).
A famous result. Block leaves at the same angle regardless of mass or radius. The combination of energy conservation and centripetal requirement uniquely determines the answer.
· two-spring system

(a) Both springs compress/extend together when block moves. Effective \(k_{\text{eff}} = k_1 + k_2 = 200\) N/m (springs in parallel). PE at displacement \(x\): \(U = \tfrac{1}{2}(200)(0.01) = 1\) J. Max KE = 1 J. Max speed: \(v = \sqrt{2K/m} = \sqrt{1} = 1\) m/s.

(b) Energy conservation: amplitude on the other side equals 0.1 m (symmetric oscillation in absence of friction).

(c) Springs in series add reciprocally: \(1/k_{\text{series}} = 1/100 + 1/200 = 3/200 \Rightarrow k_{\text{series}} = 200/3 \approx 66.7\) N/m. Then this in parallel with the left spring: \(k_{\text{eff}} = 100 + 66.7 = 166.7\) N/m.

Max speed 1 m/s; amplitude 0.1 m; new \(k_{\text{eff}} \approx 167\) N/m.
· loop with friction patch

At top of loop, minimum: \(v_{\text{top}}^2 = gR\). Energy from start (height \(h\)) to top of loop (height \(2R\)), minus energy dissipated by friction over patch of length \(L\):

\(mgh = mg(2R) + \tfrac{1}{2}mv_{\text{top}}^2 + \mu_k mg L\)

\(gh = 2gR + \tfrac{1}{2}gR + \mu_k g L = \tfrac{5}{2}gR + \mu_k gL\)

\(h_{\min} = \tfrac{5}{2}R + \mu_k L\).

\(h_{\min} = \tfrac{5}{2}R + \mu_k L\)
Without friction, \(h_{\min} = 5R/2\). The friction patch adds an extra \(\mu_k L\) to the required height, exactly compensating for the energy lost.
· bungee jumper

Let \(d\) be the maximum stretch beyond natural length. Total fall: \(20 + d\). Cord stretches by \(d\), storing PE \(\tfrac{1}{2}kd^2\). At max stretch, jumper momentarily at rest. Energy conservation: \(mg(20 + d) = \tfrac{1}{2}kd^2\).

(a) \(600(20 + d) = 25 d^2 \Rightarrow d^2 - 24d - 480 = 0 \Rightarrow d = (24 + \sqrt{576 + 1920})/2 = (24 + 49.96)/2 \approx 37\) m.

(b) Max speed during fall is when net force = 0, which is when the cord is stretched enough that spring force balances weight: \(kd^* = mg \Rightarrow d^* = 600/50 = 12\) m. At this point, fallen \(20 + 12 = 32\) m, cord stretched 12 m. Energy: \(\tfrac{1}{2}mv^2 = mg(32) - \tfrac{1}{2}kd^{*2} = 600(32) - 25(144) = 19{,}200 - 3600 = 15{,}600\) J. \(v = \sqrt{2(15600)/60} = \sqrt{520} \approx 22.8\) m/s.

(c) Free fall from same total height (32 m): \(v = \sqrt{2g(32)} = \sqrt{640} \approx 25.3\) m/s. Slightly faster than the bungee case because the cord absorbed some energy during stretching.

\(d \approx 37\) m; max speed ~22.8 m/s; about 10% slower than free fall would be at the same depth.
· escape velocity

At launch: \(K_i = \tfrac{1}{2}mv^2\), \(U_i = -GMm/R\). At infinity (just barely escaping): \(K_f = 0\), \(U_f = 0\). Conservation: \(\tfrac{1}{2}mv^2 - GMm/R = 0 \Rightarrow v_{\text{esc}} = \sqrt{2GM/R}\).

Earth: \(v_{\text{esc}} = \sqrt{2(6.67 \times 10^{-11})(5.97 \times 10^{24})/(6.37 \times 10^6)} = \sqrt{1.25 \times 10^8} \approx 11{,}200\) m/s = 11.2 km/s. ✓

\(v_{\text{esc}} = \sqrt{2GM/R}\); Earth: 11.2 km/s, matches the well-known value.
Escape velocity doesn't depend on the launching object's mass — a baseball and a rocket need the same launch speed to escape Earth's gravity. Notice it also depends only on \(M\) and \(R\) of the planet, not on \(g\) at the surface (though they're related: \(g = GM/R^2\), so \(v_{\text{esc}} = \sqrt{2gR}\), an alternate form).
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08
Common mistakes

Seven errors that cost students points on Unit 3

Forgetting the ½ in spring potential energy

A spring with \(k = 100\) N/m is compressed 0.2 m. Find its PE.

\(U_s = kx^2 = 100(0.04) = 4\) J.
\(U_s = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(100)(0.04) = 2\) J.

Spring PE comes from integrating spring force \(F = kx\) over displacement, giving \(\tfrac{1}{2}kx^2\). The "½" gets dropped most often when students confuse the spring force \(kx\) with the spring PE.

A 2 kg block on a frictionless table is attached to a spring with \(k = 50\) N/m. The block is pulled 0.4 m from equilibrium and released. Find its maximum speed.

Forgetting to include \(\cos\theta\) in work

A 50 N rope pulls a sled 10 m horizontally. The rope makes a 30° angle above horizontal. Find the work done.

\(W = Fd = (50)(10) = 500\) J.
\(W = Fd\cos\theta = (50)(10)\cos 30° = 500(0.866) \approx 433\) J. Only the horizontal component of force does work in horizontal motion.

Forces have components; only the component parallel to displacement does work. Forgetting the \(\cos\theta\) leads to overestimating work by ignoring this geometry.

A worker pushes a 30 kg crate up a 20° frictionless ramp by applying a 200 N force directed up along the ramp. The crate moves 5 m up the ramp. Find (a) the work done by the worker, (b) the work done by gravity.

Treating the normal force as doing nonzero work

A book slides 2 m across a horizontal table. The normal force from the table does how much work?

\(W = Fd = N(2) = 2N\) (something nonzero).
Zero. The normal force is perpendicular to the displacement (\(\theta = 90°\)), so \(W = N \cdot d \cos 90° = 0\).

Forces perpendicular to motion do no work, even if they're large. This is why a ball swinging in a horizontal circle keeps the same speed despite the string tension being huge — tension is perpendicular to motion.

A 5 kg block slides 10 m down a frictionless 30° incline. Find the work done by (a) gravity, (b) the normal force, (c) the net work.

Writing \(K_i + U_i = K_f + U_f\) when friction is present

A 2 kg block slides from rest down a 5 m frictionless incline at 30°, then 10 m along a horizontal surface with \(\mu_k = 0.2\). Find its final speed.

\(mgh = \tfrac{1}{2}mv^2\), so \(v = \sqrt{2gh} = \sqrt{2(10)(2.5)} = 7.07\) m/s.
Friction dissipates energy on the horizontal section. Correct equation: \(mgh = \tfrac{1}{2}mv^2 + \mu_k mg \cdot d_{\text{horiz}}\). Solve: \(50 = v^2 + 4(10) \Rightarrow v^2 = 10 \Rightarrow v \approx 3.16\) m/s.

Mechanical energy conservation only works if no non-conservative forces (friction, drag) do work. When they do, you must add the dissipated energy term \(f_k d\) to the equation.

A 1 kg block is launched from a horizontal frictionless spring (\(k = 200\) N/m, compressed 0.3 m), then slides across a rough surface (\(\mu_k = 0.3\), length 2 m) before reaching a frictionless ramp. Find its speed at the bottom of the ramp.

Confusing power with energy

A 100 W lightbulb runs for 30 minutes. How much energy does it consume?

100 W (mistakenly stating power as the energy answer).
Energy = power × time = 100(1800) = 180{,}000 J = 180 kJ. Power is the rate; energy is the total amount.

Watts measure rate of energy use; joules measure energy. A "100 W" device uses 100 J every second. The total energy depends on how long it runs.

A motor lifts a 50 kg load 20 m in 10 seconds at constant speed. Find (a) the work done, (b) the average power, (c) if the motor's efficiency is 70%, the electrical energy consumed.

Choosing the wrong reference height

A pendulum swings from height 2 m to height 0.5 m above the floor. Use the floor as reference.

Compute \(U_g\) at 2 m as \(2mg\) and at 0.5 m as \(0.5mg\) — fine, this works.
Either reference is fine — but choosing the lowest point of motion (0.5 m here) makes \(U_g\) at the lowest point zero, eliminating one term from the equation. Cleaner: \(U_i = mg(1.5)\), \(U_f = 0\).

Reference height is your choice. Choosing the lowest point as zero PE keeps all PEs positive and minimizes algebra. Don't make life harder than it needs to be.

A roller coaster starts at the top of a 30 m hill, descends to a 10 m valley, then climbs to a 20 m hill. Treating it as frictionless, find the speed at the top of the second hill if it started from rest.

Using \(v\) instead of \(v^2\) in KE

A car's speed doubles from 10 to 20 m/s. By what factor does its KE change?

It doubles.
It quadruples. \(K \propto v^2\), so doubling speed multiplies KE by 4. This is why doubling your driving speed quadruples your stopping distance under braking.

The square in kinetic energy has profound consequences for everyday physics — collisions, braking, falls. Speed up by 50%, KE goes up by \(1.5^2 = 2.25\) — over twice as much energy to dissipate in a crash.

A 1500 kg car can stop in 25 m from 30 m/s using maximum braking. Estimate the stopping distance from 60 m/s (assuming the same maximum braking force).

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09
Tutor's corner

What I learned going through this unit — and from helping others through it

What every student struggles with (in order)

The #1 sticking point in Unit 3 is recognizing when to use energy methods instead of Newton's laws. Students who learn Unit 2 deeply often try to brute-force every Unit 3 problem with FBDs and \(F = ma\), then get stuck on a curving path or a complicated force. The trigger to switch to energy methods is simple: if the question asks "how fast?", "how high?", or "how far?", and doesn't mention time or instantaneous force, energy is your tool.

The #2 struggle is the spring PE formula. The factor of ½ is non-obvious and gets dropped half the time on first try. Drill problems involving springs until \(\tfrac{1}{2}kx^2\) is automatic.

The #3 struggle is the work formula's sign and angle. Students treat work as if it's always positive — but friction does negative work, and forces perpendicular to motion do zero. Pay attention to \(\cos\theta\), and remember that 90° gives zero, 180° gives negative.

What separates a 4 from a 5

Energy bar charts. Students who score 5 can sketch energy bar charts at multiple points in any problem and instantly see the conservation principle in action. The 2025 AP Physics 1 exam featured an FRQ that asked students to draw energy bar charts at three different positions in a block-spring-ramp scenario. If you can do those charts confidently, you have the conceptual core of Unit 3 mastered. Practice them — for every energy problem you do, try drawing bar charts before and after.

The question that reveals true understanding

"A ball slides down a frictionless ramp from rest at height \(h\). At the bottom, what determines its speed?" The shallow answer: "energy conservation, so \(v = \sqrt{2gh}\)." The deep answer: "The ramp's shape doesn't matter at all — only the height drop. Path independence is the defining feature of conservative forces, and it's why energy methods work so well." Students who internalize path-independence early can tackle complex problems (curved tracks, loops, multi-part rides) by ignoring the shape entirely and just tracking energy.

Exam-day strategy

Recognize the cue: if a problem mentions a height change, a spring, or asks about final speed/height/distance — energy first. If it asks about time, instantaneous acceleration, or specific forces — Newton's laws.

Set up energy conservation as a single equation: KE + PE_grav + PE_spring (initial) = KE + PE_grav + PE_spring (final), plus a friction loss term \(f_k d\) on the right if friction is present. Then plug in. This single template covers the vast majority of Unit 3 problems.

Choose your reference height wisely: usually the lowest point in the motion, which makes that point's PE zero and keeps everything else positive.

Always check signs: positive work increases KE; negative work decreases it; friction is always negative work; gravity is positive when an object falls and negative when it rises.

A surprising real-world connection

Why are car crashes so much more dangerous at highway speeds than city speeds? Energy. A car going 60 mph has KE \( \propto v^2 = 3600\) (in arbitrary units). At 30 mph, it's 900. Doubling the speed quadruples the KE — and in a crash, all that KE has to go somewhere (deformation, heat, sound). Modern cars use crumple zones to absorb this energy gradually rather than transferring it to occupants. Same principle behind bike helmets: the foam deforms to dissipate KE over a longer distance, reducing the force on the skull. Almost every safety engineering decision involves dissipating kinetic energy — that's how Unit 3 shapes the world you live in.

One piece of advice

If you only have time to drill one type of problem, drill energy conservation with springs and gravity. The 2025 AP Physics 1 exam had a major FRQ on exactly this. The template — \(K_i + U_{g,i} + U_{s,i} = K_f + U_{g,f} + U_{s,f}\) — covers ramps, springs, free fall, pendulums, roller coasters, and most of Unit 7 (oscillations) too. Worksheet B problem 19 in this unit is built directly on the 2025 exam structure. Master it and you've covered a huge chunk of the material that's likely to show up on the exam.

End of Unit 3 · Work, Energy & Power   ·  First Principles Physics