AP Physics 1: Algebra-Based Practice Quiz: Power

Correct Answer: C

According to the provided content, power is defined as the rate at which energy changes with respect to time, either by transfer or conversion.

Correct Answer: B

Average power is calculated as the work done divided by the time interval. Using the formula $P_{avg} = \frac{W}{\Delta t}$, we get $P_{avg} = \frac{4000 \text{ J}}{20 \text{ s}} = 200 \text{ W}$.

Correct Answer: B

Average power is the change in energy divided by the change in time. Using the formula $P_{avg} = \frac{\Delta E}{\Delta t}$, we calculate $P_{avg} = \frac{500 \text{ J}}{10 \text{ s}} = 50 \text{ W}$.

Correct Answer: C

Instantaneous power is calculated as $P_{inst} = Fv\cos\theta$. Since the force is in the same direction as the velocity, the angle $\theta$ is 0°, and $\cos(0°) = 1$. Therefore, $P_{inst} = (30 \text{ N})(4 \text{ m/s})(1) = 120 \text{ W}$.

Correct Answer: A

Power for Machine X is $P_X = \frac{W}{\Delta t} = \frac{150 \text{ J}}{5 \text{ s}} = 30 \text{ W}$. Power for Machine Y is $P_Y = \frac{\Delta E}{\Delta t} = \frac{200 \text{ J}}{10 \text{ s}} = 20 \text{ W}$. Since 30 W > 20 W, Machine X has a greater average power.

Correct Answer: B

The instantaneous power is given by the formula $P_{inst} = Fv\cos\theta$. The force parallel to the velocity is $F_{||} = F\cos\theta$. Plugging in the values: $P_{inst} = (50 \text{ N})(2 \text{ m/s})\cos(30°) \approx (100)(0.87) = 87 \text{ W}$.

Correct Answer: C

The formula $P_{avg} = \frac{W}{\Delta t}$ shows that average power is the amount of work (W) done over a specific time interval ($\Delta t$), which is the definition of the rate at which work is done.

Correct Answer: A

The formula for instantaneous power is $P_{inst} = Fv\cos\theta$. The value of $\cos\theta$ is maximized when $\theta = 0°$, where $\cos(0°) = 1$. This occurs when the force is parallel to the velocity.

Correct Answer: C

To lift the elevator at a constant velocity, the motor must apply an upward force equal in magnitude to the elevator's weight. So, F = 4000 N. This force is applied in the same direction as the velocity (upward), so $\theta = 0°$. Using $P_{inst} = Fv\cos\theta$, we get $P_{inst} = (4000 \text{ N})(3 \text{ m/s})\cos(0°) = 12,000 \text{ W}$.

Correct Answer: D

Power is defined as $P = \frac{W}{\Delta t}$ and $P = Fv\cos\theta$. If the object does not move, its velocity (v) is zero. Therefore, the instantaneous power delivered is zero. Also, since there is no displacement, the work done is zero, which means the average power is also zero. The net force could be zero (if it's at rest) but the applied force itself is not necessarily zero.

Correct Answer: A

The centripetal force is always directed towards the center of the circle, while the instantaneous velocity of the object is tangent to the circle. Therefore, the angle $\theta$ between the force and velocity vectors is always 90°. Since $\cos(90°) = 0$, the instantaneous power delivered by the centripetal force is $P_{inst} = Fv\cos(90°) = 0 \text{ W}$.