AP Physics C Mechanics Quiz: Power
10 questions · exam conditions
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PowerQuestion 1 of 10

A 90 kg load is lifted 6.0 m using an ideal block-and-tackle with mechanical advantage 2, so the rope tension is T=mg2T=\tfrac{mg}{2} while the load rises at constant speed. Because two rope segments support the load, the free end of the rope must be pulled 12 m to raise the load 6.0 m. The pull takes 8.0 s, and friction is negligible. Use g=9.8m/s2g=9.8\,\text{m/s}^2. In the given system, what is the average power output of the system described?

6.6×102W6.6\times10^{2}\,\text{W}
1.3×103W1.3\times10^{3}\,\text{W}
3.3×102W3.3\times10^{2}\,\text{W}
5.5×102W5.5\times10^{2}\,\text{W}
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Power

Practice Power in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Power, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A 90 kg load is lifted 6.0 m using an ideal block-and-tackle with mechanical advantage 2, so the rope tension is T=mg2T=\tfrac{mg}{2} while the load rises at constant speed. Because two rope segments support the load, the free end of the rope must be pulled 12 m to raise the load 6.0 m. The pull takes 8.0 s, and friction is negligible. Use g=9.8m/s2g=9.8\,\text{m/s}^2. In the given system, what is the average power output of the system described?

  1. 6.6×102W6.6\times10^{2}\,\text{W} (correct answer)
  2. 1.3×103W1.3\times10^{3}\,\text{W}
  3. 3.3×102W3.3\times10^{2}\,\text{W}
  4. 5.5×102W5.5\times10^{2}\,\text{W}
Explanation: This question tests power in a mechanical advantage system. With a block-and-tackle system having mechanical advantage 2, the rope tension is half the load weight, but the rope must be pulled twice the distance. Given: mass m = 90 kg, load lift height h = 6.0 m, rope pull distance = 12 m, time t = 8.0 s, g = 9.8 m/s². The weight is W = mg = 90 × 9.8 = 882 N. The rope tension is T = mg/2 = 441 N. The work done by pulling the rope is W = T × rope distance = 441 × 12 = 5292 J. This equals the gravitational potential energy gained: mgh = 90 × 9.8 × 6.0 = 5292 J ✓. The power is P = W/t = 5292/8.0 = 661.5 W ≈ 6.6 × 10² W. Choice A is correct. The mechanical advantage doesn't change the work required (energy is conserved), but it does change the force-distance trade-off.

Question 2

A uniform disk in a lab is spun by a motor that applies a constant torque of 12 N·m while the disk rotates at a steady angular speed of 15 rad/s. Assume the angular speed is constant during the interval so the motor's torque transfers energy at a constant rate. Neglect bearing friction and air resistance so essentially all motor work becomes rotational energy transfer. Use the rotational power relation P=τωP=\tau\omega. In the given system, calculate the power exerted by the torque in the scenario.

  1. 1.8×102W1.8\times10^{2}\,\text{W} (correct answer)
  2. 2.7×101W2.7\times10^{1}\,\text{W}
  3. 8.0×101W8.0\times10^{1}\,\text{W}
  4. 1.8×103W1.8\times10^{3}\,\text{W}
Explanation: This question tests rotational power, which follows the same principles as linear power but uses rotational quantities. For rotational motion at constant angular velocity, power is given by P = τω, where τ is torque and ω is angular velocity. Given: torque τ = 12 N·m, angular velocity ω = 15 rad/s. The power is simply P = τω = 12 × 15 = 180 W = 1.8 × 10² W. Choice A is correct because it directly applies the rotational power formula. This is analogous to P = Fv for linear motion. Choice B (27 W) might result from an arithmetic error, while choice D (1800 W) would come from a decimal place error. This problem demonstrates that power concepts extend naturally to rotational systems, maintaining the same interpretation as rate of energy transfer.

Question 3

A 1500 kg elevator starts from rest and reaches 4.0 m/s in 5.0 s while moving upward, rising 10 m during that interval. The cable tension does work that increases both gravitational potential energy mghmgh and kinetic energy 12mv2\tfrac12 mv^2. Take g=9.8m/s2g=9.8\,\text{m/s}^2 and neglect frictional losses. Treat the power as the average rate of total mechanical energy increase over the 5.0 s. In the given system, what is the average power output of the system described?

  1. 3.0×104W3.0\times10^{4}\,\text{W}
  2. 2.9×104W2.9\times10^{4}\,\text{W}
  3. 3.2×104W3.2\times10^{4}\,\text{W} (correct answer)
  4. 1.5×104W1.5\times10^{4}\,\text{W}
Explanation: This question tests power calculation when both kinetic and potential energy change. Given: mass m = 1500 kg, final velocity v = 4.0 m/s (starting from rest), height h = 10 m, time t = 5.0 s, g = 9.8 m/s². The total mechanical energy increase includes both potential and kinetic energy. Potential energy increase: ΔPE = mgh = 1500 × 9.8 × 10 = 147,000 J. Kinetic energy increase: ΔKE = ½mv² = ½ × 1500 × 4.0² = ½ × 1500 × 16 = 12,000 J. Total energy increase: ΔE = ΔPE + ΔKE = 147,000 + 12,000 = 159,000 J. Average power: P = ΔE/t = 159,000/5.0 = 31,800 W ≈ 3.2 × 10⁴ W. Choice C is correct because it accounts for both forms of mechanical energy gain. This problem emphasizes that power calculations must consider all energy changes in the system.

Question 4

An electric motor hoists a 120 kg load straight up 15 m in 12 s at constant speed, so the mechanical work is mghmgh. The motor draws 2.5 kW of electrical power from the outlet while lifting, and its efficiency is 78%, meaning Pmech=ηPinP_{\text{mech}}=\eta P_{\text{in}}. Assume g=9.8m/s2g=9.8\,\text{m/s}^2 and neglect losses other than those in the efficiency rating. In the given system, electrical energy is converted into gravitational potential energy. Determine the mechanical power output given the conditions.

  1. 2.0×103W2.0\times10^{3}\,\text{W}
  2. 3.2×103W3.2\times10^{3}\,\text{W}
  3. 1.9×103W1.9\times10^{3}\,\text{W} (correct answer)
  4. 7.8×102W7.8\times10^{2}\,\text{W}
Explanation: This question tests the relationship between electrical power input, mechanical power output, and efficiency in real-world systems. Given: mass m = 120 kg, height h = 15 m, time t = 12 s, electrical power input P_in = 2.5 kW = 2500 W, efficiency η = 78% = 0.78, and g = 9.8 m/s². The mechanical power output is related to electrical input by efficiency: P_mech = η × P_in = 0.78 × 2500 = 1950 W ≈ 1.9 × 10³ W. We can verify this makes sense by checking the work required: W = mgh = 120 × 9.8 × 15 = 17,640 J, so the power needed is P = W/t = 17,640/12 = 1470 W. Since 1950 W > 1470 W, the motor has sufficient power to lift the load. Choice C is correct because it properly applies the efficiency relationship. This problem illustrates that real motors convert only a fraction of electrical power to mechanical work, with the rest lost as heat.

Question 5

An electric motor lifts a 25.0kg25.0\,\text{kg} load vertically upward at constant speed through 10.0m10.0\,\text{m} in 5.0s5.0\,\text{s}. The motor's efficiency is 80%80\%, meaning Pmech=0.80PelecP_{\text{mech}}=0.80\,P_{\text{elec}}. Ignore rotational kinetic energy of the motor and any frictional losses besides the stated efficiency. Take g=9.8m/s2g=9.8\,\text{m/s}^2 and assume the tension in the cable equals the load's weight. Based on the scenario above, determine the mechanical power output given the conditions.

  1. 392W392\,\text{W}
  2. 490W490\,\text{W} (correct answer)
  3. 613W613\,\text{W}
  4. 49.0W49.0\,\text{W}
Explanation: This question tests understanding of mechanical power and efficiency. The motor lifts a 25.0 kg load 10.0 m in 5.0 s at constant speed. The work done equals the change in gravitational potential energy: W = mgh = (25.0 kg)(9.8 m/s²)(10.0 m) = 2450 J. The mechanical power output is P = W/t = 2450 J / 5.0 s = 490 W. Choice B is correct at 490 W. The 80% efficiency relates mechanical power to electrical power (Pmech = 0.80 × Pelec), but the question asks for mechanical power output, which is 490 W. Choice A (392 W) might result from incorrectly applying the efficiency. Choice C (613 W) could be the electrical power input. Choice D (49.0 W) is off by a factor of 10.

Question 6

A flywheel is driven so that its angular speed increases uniformly from 10.0rad/s10.0\,\text{rad/s} to 30.0rad/s30.0\,\text{rad/s} in 4.0s4.0\,\text{s} while the applied torque remains constant at 8.0N0˘0b7m8.0\,\text{N\u00b7m}. Assume friction is negligible and the torque is always in the direction of rotation. The instantaneous rotational power is P=τωP=\tau\omega, so average power over the interval can be found using the average angular speed. Use SI units throughout. In the given system, what is the average power output of the system described?

  1. 80W80\,\text{W}
  2. 160W160\,\text{W} (correct answer)
  3. 320W320\,\text{W}
  4. 20W20\,\text{W}
Explanation: This problem involves calculating average rotational power during angular acceleration. With constant torque τ = 8.0 N·m and angular speeds changing from 10.0 to 30.0 rad/s, the average angular speed is ωavg = (10.0 + 30.0)/2 = 20.0 rad/s. The average power is Pavg = τ × ωavg = (8.0 N·m)(20.0 rad/s) = 160 W. Choice B is correct at 160 W. Choice A (80 W) uses the initial angular speed instead of the average. Choice C (320 W) doubles the correct answer. Choice D (20 W) appears to be a calculation error.

Question 7

A motor draws an electrical input power of 1.50kW1.50\,\text{kW} while lifting a 50.0kg50.0\,\text{kg} mass straight up at constant speed. The motor's efficiency is 70%70\%, so only 70%70\% of the electrical power becomes mechanical power delivered to the load. Neglect frictional losses other than the stated efficiency and ignore any change in kinetic energy. Use SI units and take g=9.8m/s2g=9.8\,\text{m/s}^2. Based on the scenario above, determine the mechanical power output given the conditions.

  1. 1.05kW1.05\,\text{kW} (correct answer)
  2. 1.50kW1.50\,\text{kW}
  3. 0.70kW0.70\,\text{kW}
  4. 105W105\,\text{W}
Explanation: This question tests understanding of motor efficiency and power conversion. The motor draws 1.50 kW electrical power with 70% efficiency, so the mechanical power output is Pmech = 0.70 × 1.50 kW = 1.05 kW. Choice A is correct at 1.05 kW. This mechanical power is what's available to lift the load at constant speed. Choice B (1.50 kW) is the electrical input power, not the mechanical output. Choice C (0.70 kW) might result from a calculation error. Choice D (105 W) is off by a factor of 10, likely a unit conversion error.

Question 8

Two identical motors are used to lift loads. Motor A lifts a 200 kg mass at 2.0 m/s constant velocity. Motor B lifts a 400 kg mass at constant velocity. If both motors operate at the same mechanical power output, what is the ratio of the time taken by Motor B to lift its load through 20 m compared to Motor A lifting its load through the same height?

  1. 1:1
  2. 1:2
  3. 2:1 (correct answer)
  4. 4:1
Explanation: Motor A: P_A = F_A × v_A = m_A g × v_A = 200 × 9.8 × 2.0 = 3920 W. Since both motors have the same power output, P_B = P_A = 3920 W. For Motor B: P_B = m_B g × v_B, so v_B = P_B/(m_B g) = 3920/(400 × 9.8) = 1.0 m/s. Time for Motor A to lift 20 m: t_A = 20/2.0 = 10 s. Time for Motor B to lift 20 m: t_B = 20/1.0 = 20 s. Ratio t_B : t_A = 20:10 = 2:1. Choice A (1:1) would result from thinking both motors take the same time regardless of load. Choice B (1:2) would result from inverting the ratio. Choice D (4:1) might result from using the mass ratio instead of considering the velocity relationship.

Question 9

An electric motor operates at constant mechanical power output of 1.5 kW while lifting a load vertically. Initially, the load has a mass of 100 kg and is lifted at constant velocity. After 10 seconds, an additional 50 kg mass is suddenly attached to the load. Assuming the motor maintains the same power output, what is the new constant velocity of the combined load after the transient effects have died down?

  1. 1.0 m/s (correct answer)
  2. 1.2 m/s
  3. 1.5 m/s
  4. 2.0 m/s
Explanation: Initially, with m₁ = 100 kg at constant velocity v₁: P = F₁v₁ = m₁gv₁, so v₁ = P/(m₁g) = 1500/(100×9.8) = 1.53 m/s. After the additional mass is added, the total mass is m₂ = 150 kg. At the new constant velocity v₂: P = F₂v₂ = m₂gv₂, so v₂ = P/(m₂g) = 1500/(150×9.8) = 1.02 m/s ≈ 1.0 m/s. Choice B (1.2 m/s) might result from using an incorrect mass ratio or calculation error. Choice C (1.5 m/s) would result from not accounting for the increased mass properly. Choice D (2.0 m/s) would result from incorrectly thinking the velocity increases when mass increases.

Question 10

A wind turbine generates electrical power according to P=0.4v3P = 0.4v^3 kW, where v is the wind speed in m/s. During a particular day, the wind speed varies as v(t)=8+2sin(πt/12)v(t) = 8 + 2\sin(\pi t/12) m/s, where t is time in hours. What is the average power generated during the first 12 hours?

  1. 205 kW
  2. 220 kW (correct answer)
  3. 235 kW
  4. 250 kW
Explanation: The power is P(t) = 0.4[v(t)]³ = 0.4[8 + 2sin(πt/12)]³. To find the average power over 12 hours: P_avg = (1/12)∫₀¹² 0.4[8 + 2sin(πt/12)]³ dt. Let u = πt/12, so du = π dt/12, and dt = 12du/π. When t = 0, u = 0; when t = 12, u = π. The integral becomes: P_avg = (1/12) × 0.4 × (12/π) ∫₀^π [8 + 2sin(u)]³ du = (0.4/π) ∫₀^π [8 + 2sin(u)]³ du. Expanding: [8 + 2sin(u)]³ = 512 + 384sin(u) + 96sin²(u) + 8sin³(u). Using standard integrals: ∫₀^π sin(u)du = 0, ∫₀^π sin²(u)du = π/2, ∫₀^π sin³(u)du = 0. So: ∫₀^π [8 + 2sin(u)]³ du = 512π + 0 + 96(π/2) + 0 = 512π + 48π = 560π. Therefore: P_avg = (0.4/π) × 560π = 224 kW ≈ 220 kW.