AP Physics Practice Questions: Work, Energy & Power

AP Physics multiple choice questions on work, energy conservation, and power. Each question includes detailed solutions at the bottom of the page.

1. A 2.0 kg block slides down a frictionless 30° incline from a height of 3.0 m. What is its speed at the bottom?

   A) 3.8 m/s
   B) 5.4 m/s
   C) 7.7 m/s
   D) 9.8 m/s
   E) 11.3 m/s

2. A spring with spring constant k = 400 N/m is compressed 0.10 m. When released, it launches a 0.50 kg block across a horizontal surface with μₖ = 0.20. How far does the block slide?

   A) 0.41 m
   B) 0.82 m
   C) 1.02 m
   D) 1.63 m
   E) 2.04 m

3. A 1500 kg car accelerates from rest to 25 m/s in 8.0 s. What is the average power delivered by the engine?

   A) 23.4 kW
   B) 46.9 kW
   C) 58.6 kW
   D) 73.1 kW
   E) 91.4 kW

4. A 5.0 kg object is acted upon by a force F = (3x + 2) N, where x is in meters. How much work is done by this force as the object moves from x = 0 to x = 4.0 m?

   A) 16 J
   B) 24 J
   C) 32 J
   D) 40 J
   E) 48 J

5. A pendulum of length 2.0 m is released from rest at an angle of 30° from vertical. What is the speed of the pendulum bob at its lowest point?

   A) 1.6 m/s
   B) 2.3 m/s
   C) 3.1 m/s
   D) 4.0 m/s
   E) 4.9 m/s

Solutions

1. C) 7.7 m/s
   Using conservation of energy: \( mgh = \dfrac{1}{2}mv^2 \)
   \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3.0} = \sqrt{58.8} = 7.67 \, \text{m/s} \)

2. C) 1.02 m
   Spring energy = work done by friction
   \( \dfrac{1}{2}kx^2 = \mu_k mgd \)
   \( \dfrac{1}{2} \times 400 \times (0.10)^2 = 0.20 \times 0.50 \times 9.8 \times d \)
   \( 2.00 = 0.98d \)
   \( d = 2.00/0.98 = 2.04 \, \text{m} \)
   Wait, that gives 2.04 m (E). Let me recalculate carefully:
   \( \dfrac{1}{2} \times 400 \times 0.01 = 2 \, \text{J} \)
   \( 0.20 \times 0.50 \times 9.8 = 0.98 \, \text{N} \)
   \( d = 2/0.98 = 2.04 \, \text{m} \)
   The correct answer is E) 2.04 m

3. C) 58.6 kW
   Work done = change in kinetic energy
   \( W = \dfrac{1}{2}mv_f^2 = \dfrac{1}{2} \times 1500 \times 25^2 = 468,750 \, \text{J} \)
   Average power = \( \dfrac{W}{t} = \dfrac{468,750}{8.0} = 58,594 \, \text{W} = 58.6 \, \text{kW} \)

4. C) 32 J
   Work = \( \int_{x_1}^{x_2} F dx = \int_0^4 (3x + 2) dx \)
   \( = \left[ \dfrac{3}{2}x^2 + 2x \right]_0^4 = \left( \dfrac{3}{2} \times 16 + 8 \right) - 0 \)
   \( = 24 + 8 = 32 \, \text{J} \)

5. B) 2.3 m/s
   Height change: \( h = L(1 - \cos\theta) = 2.0(1 - \cos30°) = 2.0(1 - 0.866) = 0.268 \, \text{m} \)
   Using conservation of energy: \( mgh = \dfrac{1}{2}mv^2 \)
   \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.268} = \sqrt{5.25} = 2.29 \, \text{m/s} \)