Free MCAT-style practice, 4A
MCAT Work, Energy and Power Practice Questions
Ten free MCAT-style practice questions on work, the work-energy theorem, gravitational and spring potential energy, energy conservation, power and simple machines, with full explanations.
Quick concept primer
Work is the energy a force transfers along a displacement: W = Fd cos θ, where θ is the angle between the force and the displacement. A force perpendicular to the motion (a normal force on level ground, the tension that keeps a ball in uniform circular motion) does zero work. The work-energy theorem says the net work done by all forces equals the change in kinetic energy, ΔKE = ½mv2final − ½mv2initial.
Gravity and ideal springs are conservative forces: the work they do depends only on the start and end points, so it can be stored as potential energy, mgh for gravity and ½kx2 for a spring. When only conservative forces do work, KE + PE stays constant. Friction is nonconservative: the mechanical energy it removes appears as thermal energy. Power is the rate of doing work, P = W/t = Fv, measured in watts. An ideal simple machine (lever, pulley, ramp) lowers the force you need by making you apply it over a longer distance; the work stays the same.
Traps: work uses only the force component along the displacement, a machine never reduces the work required, and doubling a spring's compression quadruples its stored energy. These questions use g = 10 m/s2.
10 practice questions
Question 1
A rope pulls a crate 5.0 m across a level floor. The rope tension is 50 N and the rope makes a 37 degree angle above the horizontal (cos 37 degrees = 0.80, sin 37 degrees = 0.60). How much work does the tension do on the crate?
- 150 J
- 200 J
- 250 J
- 1000 J
Show answer and explanation
Answer: B. 200 J
Only the component of the tension along the displacement does work: W = Fd cos θ = 50 x 5.0 x 0.80 = 200 J. Using sin 37 degrees gives the 150 J distractor, and ignoring the angle gives 250 J. The vertical component of the tension does no work because the crate does not move vertically.
Question 2
A stone is thrown straight down at 15 m/s from a window 10 m above the ground. Ignoring air resistance and using g = 10 m/s2, what is its speed just before it hits the ground?
- 14 m/s
- 15 m/s
- 18 m/s
- 21 m/s
Show answer and explanation
Answer: D. 21 m/s
Conservation of energy gives v2 = v02 + 2gh = 225 + 200 = 425, so v is about 20.6 m/s, about 21 m/s. The 14 m/s choice is the speed a stone dropped from rest would reach, which ignores the initial kinetic energy. The answer would be the same if the stone were thrown upward or sideways at 15 m/s, because energy is a scalar.
Question 3
A 2.0 kg cart moving at 2.0 m/s on a level track is pushed forward through a 3.0 m stretch. The forces acting during that stretch are listed below.
| Force | Magnitude | Direction |
|---|---|---|
| Applied push | 12 N | forward |
| Kinetic friction | 4 N | backward |
| Normal force | 20 N | up |
| Gravity | 20 N | down |
What is the cart's speed at the end of the 3.0 m stretch?
- 2.8 m/s
- 4.0 m/s
- 4.9 m/s
- 5.3 m/s
Show answer and explanation
Answer: D. 5.3 m/s
The normal force and gravity are perpendicular to the motion and do no work. Net work = (12 − 4) x 3.0 = 24 J. Initial KE = ½(2.0)(2.0)2 = 4 J, so final KE = 28 J and v = √(2 x 28 / 2.0) = √28, about 5.3 m/s. The 4.9 m/s choice forgets the initial kinetic energy.
Question 4
A spring with spring constant 400 N/m is compressed 0.10 m and released, launching a 0.050 kg block across a frictionless horizontal surface. What is the block's launch speed?
- 4.0 m/s
- 8.9 m/s
- 13 m/s
- 80 m/s
Show answer and explanation
Answer: B. 8.9 m/s
The stored energy is ½kx2 = ½ x 400 x (0.10)2 = 2.0 J. Setting 2.0 J = ½(0.050)v2 gives v2 = 80 and v of about 8.9 m/s. The 80 m/s choice reports v2 instead of v. Forgetting to square the compression would give a much larger, incorrect energy.
Question 5
A 70 kg student climbs stairs that raise her 3.0 m vertically in 4.2 s at constant speed. What average power does she supply against gravity?
- 50 W
- 210 W
- 500 W
- 2100 W
Show answer and explanation
Answer: C. 500 W
The work against gravity is mgh = 70 x 10 x 3.0 = 2100 J, and power is work divided by time: 2100 J / 4.2 s = 500 W. The 2100 W choice forgets to divide by time, and 50 W leaves out g. Constant speed means no change in kinetic energy, so all the work goes into potential energy.
Question 6
An ideal, frictionless pulley system with a mechanical advantage of 4 is used to lift a 200 N load slowly through a height of 0.50 m. All of the following statements are true EXCEPT:
- The required input force is 50 N
- The rope must be pulled through 2.0 m
- The gravitational potential energy of the load rises by 100 J
- The work done by the person is 25 J
Show answer and explanation
Answer: D. The work done by the person is 25 J
An ideal machine conserves work. The person pulls with 200 / 4 = 50 N through 4 x 0.50 = 2.0 m, doing 50 x 2.0 = 100 J, exactly the 100 J gain in the load's potential energy. Choice D wrongly divides the work by the mechanical advantage; machines trade force for distance but never reduce the work.
Question 7
A box is pushed from the floor up to a shelf 1.5 m higher along two different rough ramps: one short and steep, the other long and curved. The box starts and ends at rest both times. Which quantity is the same for both routes?
- The work done by gravity on the box
- The work done by kinetic friction on the box
- The thermal energy produced at the ramp surface
- The total work done by the person pushing the box
Show answer and explanation
Answer: A. The work done by gravity on the box
Gravity is conservative, so its work depends only on the change in height: −mg(1.5 m) on either route. Friction is nonconservative and path dependent; the longer ramp generally dissipates a different amount of energy as heat, so the work by friction, the thermal energy and the person's total work all differ between routes.
Question 8
Which of the following forces does zero work on the object it acts on?
- The tension in a string that keeps a ball moving in a horizontal circle at constant speed
- Kinetic friction acting on a book sliding to a stop
- Gravity acting on an apple falling from a tree
- The upward force a person applies while lifting a box onto a table
Show answer and explanation
Answer: A. The tension in a string that keeps a ball moving in a horizontal circle at constant speed
In uniform circular motion the tension points toward the center, perpendicular to the velocity at every instant, so cos 90 degrees = 0 and the tension does no work; the ball's speed and kinetic energy stay constant. Friction on the sliding book does negative work, while gravity on the falling apple and the lifting force on the box both do positive work.
Question 9
A 2.0 kg sled starts from rest at the top of a snowy hill 5.0 m high and reaches the bottom moving at 8.0 m/s. How much mechanical energy was converted to thermal energy by friction and air resistance?
- 36 J
- 64 J
- 100 J
- 164 J
Show answer and explanation
Answer: A. 36 J
The starting potential energy is mgh = 2.0 x 10 x 5.0 = 100 J. The kinetic energy at the bottom is ½(2.0)(8.0)2 = 64 J. The missing 100 − 64 = 36 J was dissipated by nonconservative forces. The 64 J and 100 J choices are the final KE and initial PE themselves, not the loss.
Question 10
A variable force acts on an object along its direction of motion. The force at several positions is listed below; between listed points it changes linearly.
| Position (m) | Force (N) |
|---|---|
| 0 | 10 |
| 2 | 10 |
| 4 | 30 |
How much work does the force do as the object moves from 0 to 4 m?
- 40 J
- 50 J
- 60 J
- 80 J
Show answer and explanation
Answer: C. 60 J
Work is the area under a force versus position graph. From 0 to 2 m the area is 10 x 2 = 20 J. From 2 to 4 m the force rises linearly, so the area is the average force, (10 + 30) / 2 = 20 N, times 2 m = 40 J. The total is 60 J. Using 10 N throughout gives 40 J, and using 20 N times 4 m gives 80 J.