AP Physics 1 Work, Energy, and Power — Worked Answer Explanations

Unit 3 · 18% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Physics 1 Work, Energy, and Power practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Work, Energy, and Power practice test and come back here to review, or head back to the Work, Energy, and Power unit overview.

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Question 1 · Easy
A horizontal force of $20 N$ pushes a box $5.0 m$ across a level floor. How much work does the force do on the box?
- A
  $4.0 J$
  Why not A: Divided force by distance.
- B
  $25 J$
  Why not B: Added rather than multiplied.
- C
  $100 J$ Correct
- D
  $2000 J$
  Why not D: Off by a factor of 20, perhaps multiplied by $g$ .
Explanation
$W = F d cos θ = 20 \times 5.0 \times cos 0° = 100 J$ .

Key takeaway
Work is force times displacement along the force direction.
Question 2 · Easy
A $2.0 kg$ object's speed increases from $3.0 m/s$ to $5.0 m/s$ . What is the net work done on the object?
- A
  $4.0 J$
  Why not A: Subtracted speeds linearly and applied $\frac{1}{2} m v$ .
- B
  $8.0 J$
  Why not B: Forgot the factor of $\frac{1}{2}$ .
- C
  $16 J$ Correct
- D
  $25 J$
  Why not D: Used $\frac{1}{2} m v_{f}^{2}$ only.
Explanation
Work-energy theorem: $W_{n e t} = Δ K E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} (2) (25 - 9) = 16 J$ .

Key takeaway
$W_{net} = \Delta KE$; use the difference of squared speeds, not the difference of speeds.
Question 3 · Easy
A $0.50 kg$ ball is dropped from rest from a height of $20 m$ . Ignoring air resistance and taking $g = 10 m/s^{2}$ , what is its speed just before hitting the ground?
- A
  $10 m/s$
  Why not A: Forgot the factor of 2 inside the square root.
- B
  $14 m/s$
  Why not B: Misapplied $2 g h$ .
- C
  $20 m/s$ Correct
- D
  $200 m/s$
  Why not D: Forgot the square root entirely.
Explanation
Conservation of energy: $m g h = \frac{1}{2} m v^{2} \Rightarrow v = 2 g h = 2 \cdot 10 \cdot 20 = 20 m/s$ .

Key takeaway
Free-fall speed from height $h$ is $v = \sqrt{2gh}$, independent of mass.
Question 4 · Easy
A spring with spring constant $k = 200 N/m$ is compressed $0.10 m$ from its natural length. How much potential energy is stored in the spring?
- A
  $1.0 J$ Correct
- B
  $2.0 J$
  Why not B: Forgot the factor of $\frac{1}{2}$ .
- C
  $10 J$
  Why not C: Forgot to square the displacement.
- D
  $20 J$
  Why not D: Used $k x$ as the energy.
Explanation
$U_{s p r in g} = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ .

Key takeaway
Spring PE scales with the square of the displacement: $\tfrac{1}{2}kx^2$.
Question 5 · Easy
Which of the following statements about work is correct?
- A
  A force perpendicular to displacement does no work.Correct
- B
  Work is always positive.
  Why not B: Negative work occurs whenever force opposes motion.
- C
  Static friction can never do work.
  Why not C: On the driving wheel of a car, static friction does positive work on the car.
- D
  Work has units of newton-seconds.
  Why not D: Newton-seconds is the unit of impulse.
Explanation
Since $W = F d cos θ$ , when $θ = 90°$ (force perpendicular to displacement), $cos 90° = 0$ , so $W = 0$ . The classic example: gravity does no work on a horizontally moving object on a level surface.

Key takeaway
Only the component of force along the displacement does work.
Question 6 · Medium
A $1000 kg$ elevator rises at a constant $2.0 m/s$ . Taking $g = 10 m/s^{2}$ , what is the minimum power the motor must deliver?
- A
  $2000 W$
  Why not A: Forgot to multiply by $g$ .
- B
  $5000 W$
  Why not B: Off by a factor; possibly used $v^{2}$ .
- C
  $10, 000 W$
  Why not C: Used the weight only without multiplying by speed.
- D
  $20, 000 W$ Correct
Explanation
At constant speed, lift force equals weight $m g = 10, 000 N$ . Power $P = F v = 10, 000 \times 2.0 = 20, 000 W$ .

Key takeaway
Power delivered by a constant force moving at speed $v$ is $P = Fv$.
Question 7 · Medium
A block of mass $m$ slides down a frictionless ramp from height $h$ . At the bottom it enters a horizontal surface with kinetic friction coefficient $μ$ . How far along the horizontal surface does the block slide before stopping?
- A
  $\frac{h}{μ}$ Correct
- B
  $μ h$
  Why not B: Inverted $μ$ in the energy equation.
- C
  $\frac{h}{μ}$
  Why not C: Treated the energy equation like a kinematics one.
- D
  Depends on $m$ .
  Why not D: Mass cancels — the friction force scales with mass too.
Explanation
Energy lost to friction equals initial PE: $μ m g d = m g h$ , giving $d = h / μ$ . Mass cancels.

Key takeaway
On a friction surface, energy dissipated $= \mu m g d$; equate to PE lost.
Question 8 · Medium
A $500 kg$ roller-coaster car at the top of a $40 m$ hill is moving at $5.0 m/s$ . With $g = 10 m/s^{2}$ and ignoring friction, what is its speed at the bottom of the hill?
- A
  $25 m/s$
  Why not A: Used $2 g h$ alone, ignoring initial KE.
- B
  $28 m/s$
  Why not B: Added speeds linearly: $5 + 2 g h \approx 5 + 28$ .
- C
  $28.7 m/s$ Correct
- D
  $30 m/s$
  Why not D: Rounded $825$ incorrectly.
Explanation
Conservation: $\frac{1}{2} m v_{f}^{2} = \frac{1}{2} m v_{i}^{2} + m g h$ . So $v_{f} = v_{i}^{2} + 2 g h = 25 + 800 = 825 \approx 28.7 m/s$ .

Key takeaway
When the object starts with KE, $v_f = \sqrt{v_i^2 + 2gh}$ — speeds don't add linearly.