Work, Energy, and Power
Unit 3 of AP Physics 1, worth 18–23% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Translational kinetic energy, work, potential energy, conservation of energy, power.
How this unit is tested
Start by identifying what kind of energy quantity the problem is really asking about: motion (kinetic energy), stored energy (gravitational or elastic potential energy), a force acting over a distance (work), or a rate of energy transfer (power). Most AP Physics 1 energy problems are really disguised bookkeeping problems — you are tracking where energy starts, where it ends up, and what left the system as heat.
Before writing any equation, draw the situation and list every force acting on the object. Classify each force as conservative (gravity, ideal springs) or non-conservative (friction, applied pushes, air resistance, normal forces that do no work at all). This classification tells you whether total mechanical energy is conserved or whether you need to add a work term for the non-conservative forces.
For work calculations, find the angle between the force and the displacement — $W=Fd\cos\theta$ — and remember that any force perpendicular to the motion (normal force, centripetal tension) does zero work. If the force changes with position, work is the area under a force-versus-position graph instead of a single multiplication.
For conservation-of-energy problems, set up an energy equation comparing an initial state and a final state: $KE_i+PE_i+W_{nc}=KE_f+PE_f$. Choose a convenient reference height for potential energy (only changes in height matter), and let $W_{nc}$ be negative when friction or drag removes energy from the system.
Finally, for power, decide whether the question wants an average rate over an interval ($P=W/\Delta t$) or an instantaneous rate at one moment ($P=Fv\cos\theta$). Check that your final units make sense — joules for energy, watts for power — and that signs reflect the physical direction of energy flow.
What you have to know
Translational Kinetic Energy
$KE=\frac{1}{2}mv^2$, a scalar that depends on mass and the square of speed.
Work done by a force
For a constant force, $W=Fd\cos\theta$, where $\theta$ is the angle between force and displacement. For a varying force, $W$ equals the area under a force-versus-position graph.
Work-Energy Theorem
The net work done on an object equals its change in kinetic energy: $W_{net}=\Delta KE=KE_f-KE_i$.
Potential Energy Formulas
Gravitational (near Earth's surface): $\Delta PE_{grav}=mg\Delta h$. Elastic (ideal spring): $PE_{spring}=\frac{1}{2}kx^2$, where $x$ is displacement from equilibrium.
Conservation of Energy
If only conservative forces do work, mechanical energy is conserved: $KE_i+PE_i=KE_f+PE_f$. If non-conservative forces (like friction) also act, $KE_i+PE_i+W_{nc}=KE_f+PE_f$, where $W_{nc}$ is negative for energy-removing forces.
Power
Average power is the rate of energy transfer, $P_{avg}=\frac{W}{\Delta t}$. Instantaneous power delivered by a force is $P=Fv\cos\theta$.
14 practice questions
-
A 0.50 kg ball moves at 12 m/s. What is its translational kinetic energy?
Show the answer
Answer. 36 J
$KE=\frac{1}{2}mv^2=0.5\times0.50\times12^2=0.5\times0.50\times144=36\ J$. Kinetic energy scales with the square of speed, so doubling the speed would quadruple the kinetic energy. -
An object moves in a horizontal circle at constant speed, held by a string that provides the centripetal force. Which force does zero work on the object?
Show the answer
Answer. The tension providing centripetal force
Work depends on the component of force along the displacement, $W=Fd\cos\theta$. The tension always points toward the center, perpendicular to the object's velocity at every instant, so $\cos\theta=0$ and it does zero work — consistent with the object's constant speed and KE. -
A net force does 150 J of work on a 3.0 kg object that starts at rest. Find its final speed.
Show the answer
Answer. 10 m/s
By the work-energy theorem, $W_{net}=\Delta KE=\frac{1}{2}mv_f^2-0$. So $150=\frac{1}{2}(3.0)v_f^2$, giving $v_f^2=100$ and $v_f=10\ m/s$. -
A box slides across a horizontal frictionless floor. Which statement about the normal force acting on the box is correct?
Show the answer
Answer. It does zero work because it is perpendicular to the displacement
Since $W=Fd\cos\theta$ and the normal force is perpendicular to the horizontal displacement, $\theta=90°$ and $\cos\theta=0$, so the normal force contributes zero work regardless of its magnitude. -
A spring with k = 200 N/m is compressed 0.15 m from equilibrium. How much elastic potential energy is stored?
Show the answer
Answer. 2.25 J
$PE_{spring}=\frac{1}{2}kx^2=0.5\times200\times0.15^2=0.5\times200\times0.0225=2.25\ J$. -
A ball is thrown straight up at 20 m/s. Using energy conservation and ignoring air resistance, find the maximum height reached.
Show the answer
Answer. about 20.4 m
At maximum height all kinetic energy has converted to gravitational potential energy: $\frac{1}{2}mv^2=mgh$, so $h=\frac{v^2}{2g}=\frac{400}{19.6}\approx20.4\ m$. Mass cancels out entirely. -
A roller coaster car crests a frictionless hill, gaining height while losing speed. What happens to the total mechanical energy of the car as it goes over the hill?
Show the answer
Answer. It stays constant because only gravity, a conservative force, does work
With no friction acting, gravity is the only force doing work, and gravity is conservative. So kinetic energy converts into potential energy and back, but their sum — mechanical energy — remains constant. -
A 4.0 kg box slides down a frictionless ramp from a height of 3.0 m onto a rough horizontal floor with μ_k = 0.25. How far does the box slide across the floor before stopping?
Show the answer
Answer. 12 m
Energy entering the floor equals $mgh=4.0\times9.8\times3.0=117.6\ J$. Friction force is $f=\mu_k mg=0.25\times4.0\times9.8=9.8\ N$. Setting friction's work equal to that KE: $d=\frac{117.6}{9.8}=12\ m$. -
A motor lifts a 50 kg load 10 m straight up at constant velocity in 8.0 s. Find the motor's average power output.
Show the answer
Answer. about 613 W
At constant velocity the motor's force balances gravity, so work done is $W=mgh=50\times9.8\times10=4900\ J$. Average power is $P=W/t=4900/8.0\approx613\ W$. -
Explain why a potential energy function can be defined for gravity but not for kinetic friction.
Show the answer
Answer. Gravity's work is path-independent, depending only on start and end height, so a PE can be assigned to each position. Friction's work depends on the actual path length traveled, so no unique PE exists for it — its energy is dissipated as heat instead.
A conservative force does the same work between two points no matter which path is taken, which is exactly the property needed to define a potential energy as a function of position alone. Friction's work depends on how far the object actually slides, so it isn't a function of position only — it can't be stored and recovered, only dissipated. -
A car's engine exerts a forward force of 2000 N while the car travels at a constant 25 m/s. Find the instantaneous power output of the engine.
Show the answer
Answer. 50,000 W (50 kW)
Instantaneous power is $P=Fv\cos\theta$. Since the force acts in the direction of motion, $\theta=0°$, so $P=Fv=2000\times25=50{,}000\ W$. -
A force applied to a cart varies with position, shown on a force-versus-position graph as a straight line rising from 0 N at x = 0 to 40 N at x = 5.0 m. Find the work done on the cart over this interval.
Show the answer
Answer. 100 J
For a varying force, work equals the area under the F-x graph. This region is a triangle: $W=\frac{1}{2}(5.0\ m)(40\ N)=100\ J$. -
A pendulum bob is released from rest at height h above its lowest point. Ignoring air resistance, what is true about its energy at the lowest point?
Show the answer
Answer. All initial PE converted to KE, KE = mgh
The string tension does zero work since it is always perpendicular to the bob's velocity, and there's no friction, so mechanical energy is conserved. All the gravitational PE lost converts directly into kinetic energy: $KE=mgh$. -
A 1.0 kg block is pushed against a spring (k = 500 N/m), compressing it 0.20 m, then released. It slides across a frictionless floor and up a frictionless ramp. Find the maximum height it reaches on the ramp.
Show the answer
Answer. about 1.02 m
All spring PE converts to gravitational PE at maximum height (KE = 0 there, no friction to remove energy). $PE_{spring}=\frac{1}{2}kx^2=0.5\times500\times0.20^2=10\ J$. Setting $mgh=10\ J$ gives $h=\frac{10}{1.0\times9.8}\approx1.02\ m$.
What people get wrong
- Forgetting the $\cos\theta$ factor when a force isn't parallel to displacement, treating $W=Fd$ even when the force is applied at an angle. Always identify the angle between the force vector and the displacement vector first.
- Assuming forces like the normal force or the tension in circular motion do work simply because they are large. If a force is perpendicular to the object's displacement at every instant, it does exactly zero work.
- Using the length traveled along an incline instead of the vertical height change when computing gravitational potential energy. Only $\Delta h$ (vertical) matters for $\Delta PE_{grav}$, not the path length.
- Applying $KE_i+PE_i=KE_f+PE_f$ when friction or another non-conservative force is present. Check the force list first; if friction acts, add a $W_{nc}$ term or the answer will overestimate final speed or height.
- Confusing average power with instantaneous power, using $P=W/t$ when the problem actually asks for power at one specific moment. Use $P=Fv\cos\theta$ for an instant, and $P=W/\Delta t$ only for a rate over a time interval.
- Dropping the sign on work done by an opposing force, so the energy equation comes out too high. Work done by a force that opposes displacement is negative and must be subtracted, not added.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.