MCAT · Chemical and Physical Foundations · Physics

Work, energy and momentum

Work, energy and momentum is part of Physics in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include mechanical energy conservation on a track, work-energy theorem with friction, one-dimensional collisions and impulse and average force.

Common traps: making speed depend on mass; using path length instead of vertical height; doubling the height and doubling the speed instead of scaling by the square root.

Sample questions

Question 1
A person pushes a crate with a constant 50 N horizontal force, moving it 10 m in the direction of the force in 5 s. What average power does the person deliver?
- A.
  10 W
- B.
  25 W
- C.
  100 W
- D.
  500 W
Answer: C. 100 W
Work W = Fd = 50·10 = 500 J, and power P = W/t = 500/5 = 100 W. 500 J is the work, not the rate of doing it.
Question 2
A ball is released from rest 80 m above the ground. Ignoring air resistance and using g = 10 m/s², how fast is it moving just before it lands?
- A.
  20 m/s
- B.
  28.284 m/s
- C.
  40 m/s
- D.
  1,600 m/s
Answer: C. 40 m/s
Energy conservation: mgh = ½mv², so v = √(2gh) = √(2·10·80) = √1600 = 40 m/s. Forgetting the square root gives 1600, and dropping the 2 gives √(gh).
Question 3
The tension in a rehabilitation resistance band rises uniformly from 0 to 60 N as it is stretched 0.40 m; a stop then holds the tension at 60 N for a further 0.20 m of pull. How much work does the patient do on the band over the full 0.60 m?
- A.
  12 J
- B.
  18 J
- C.
  24 J
- D.
  36 J
Answer: C. 24 J
Work is the area under the force–displacement graph: a triangle ½(60)(0.40) = 12 J plus a rectangle (60)(0.20) = 12 J, for 24 J (C). 12 J (A) includes only the triangle; 18 J (B) treats the whole graph as one triangle; 36 J (D) multiplies the peak force by the full distance.
Question 4
An instrument case sliding across a hospital floor at 3.0 m/s comes to rest under kinetic friction with μk = 0.25. How far does it slide? (g = 10 m/s²)
- A.
  0.60 m
- B.
  1.8 m
- C.
  3.6 m
- D.
  18 m
Answer: B. 1.8 m
Friction's negative work removes the kinetic energy: μk mg d = ½mv², so d = v²/(2μk g) = 9.0/5.0 = 1.8 m (B). 0.60 m (A) uses v instead of v², making distance linear in speed; 3.6 m (C) drops the ½ in kinetic energy; 18 m (D) leaves g out of the friction force.
Question 5
On a level air track, a 1.0 kg glider moving at 3.0 m/s collides elastically and head-on with a 2.0 kg glider at rest. What is the speed of the 2.0 kg glider after the collision?
- A.
  1.0 m/s
- B.
  2.0 m/s
- C.
  3.0 m/s
- D.
  4.0 m/s
Answer: B. 2.0 m/s
For an elastic collision with the target at rest, v₂ = 2m₁v₁/(m₁ + m₂) = 2(1.0)(3.0)/3.0 = 2.0 m/s (B); the 1.0 kg glider rebounds at 1.0 m/s, and momentum and kinetic energy both check. 1.0 m/s (A) is the common speed for a perfectly inelastic collision; 3.0 m/s (C) assumes the speeds simply exchange, which happens only for equal masses; 4.0 m/s (D) puts the wrong mass in the numerator.

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