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AP Physics C: Mechanics › Power
A 90 kg load is lifted 6.0 m using an ideal block-and-tackle with mechanical advantage 2, so the rope tension is $T=\tfrac{mg}{2}$ 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 $g=9.8,\text{m/s}^2$. In the given system, what is the average power output of the system described?
$6.6\times10^{2},\text{W}$
$1.3\times10^{3},\text{W}$
$5.5\times10^{2},\text{W}$
$3.3\times10^{2},\text{W}$
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.
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 $P=\tau\omega$. In the given system, calculate the power exerted by the torque in the scenario.
$8.0\times10^{1},\text{W}$
$2.7\times10^{1},\text{W}$
$1.8\times10^{2},\text{W}$
$1.8\times10^{3},\text{W}$
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.
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 $mgh$ and kinetic energy $\tfrac12 mv^2$. Take $g=9.8,\text{m/s}^2$ 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?
$1.5\times10^{4},\text{W}$
$2.9\times10^{4},\text{W}$
$3.2\times10^{4},\text{W}$
$3.0\times10^{4},\text{W}$
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.
An electric motor hoists a 120 kg load straight up 15 m in 12 s at constant speed, so the mechanical work is $mgh$. The motor draws 2.5 kW of electrical power from the outlet while lifting, and its efficiency is 78%, meaning $P_{\text{mech}}=\eta P_{\text{in}}$. Assume $g=9.8,\text{m/s}^2$ 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.
$7.8\times10^{2},\text{W}$
$2.0\times10^{3},\text{W}$
$1.9\times10^{3},\text{W}$
$3.2\times10^{3},\text{W}$
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.
An electric motor lifts a $25.0,\text{kg}$ load vertically upward at constant speed through $10.0,\text{m}$ in $5.0,\text{s}$. The motor’s efficiency is $80%$, meaning $P_{\text{mech}}=0.80,P_{\text{elec}}$. Ignore rotational kinetic energy of the motor and any frictional losses besides the stated efficiency. Take $g=9.8,\text{m/s}^2$ 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.
$613,\text{W}$
$49.0,\text{W}$
$490,\text{W}$
$392,\text{W}$
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.
A flywheel is driven so that its angular speed increases uniformly from $10.0,\text{rad/s}$ to $30.0,\text{rad/s}$ in $4.0,\text{s}$ while the applied torque remains constant at $8.0,\text{N\u00b7m}$. Assume friction is negligible and the torque is always in the direction of rotation. The instantaneous rotational power is $P=\tau\omega$, 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?
$320,\text{W}$
$20,\text{W}$
$160,\text{W}$
$80,\text{W}$
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.
A motor draws an electrical input power of $1.50,\text{kW}$ while lifting a $50.0,\text{kg}$ mass straight up at constant speed. The motor’s efficiency is $70%$, so only $70%$ 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 $g=9.8,\text{m/s}^2$. Based on the scenario above, determine the mechanical power output given the conditions.
$105,\text{W}$
$1.50,\text{kW}$
$1.05,\text{kW}$
$0.70,\text{kW}$
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.