AP Physics 1 - Power | Practice Hub

If three locomotives are pulling a train, how much power does each locomotive need to apply on average during the first second to accelerate the train at $1\frac{cm}{s^2}$ from rest?

Explanation

Using

$v_f=.01\frac{m}{s^2}$

Using

$P=\frac{W}{t}$

and

All energy will be kinetic. Combining equations:

$P=\frac{.5mv^2_f-.5mv_i^2}{t}$

Plugging in values:

$P=\frac{.5*3000000*.01*^2-.5*3000000*0^2}{1}$

Jennifer throws a ball into the air. Her throw took . Determine the power of her throwing arm.

Explanation

Calculating gravitational potential energy:

Converting to and plugging in values:

Based on conservation of energy, this would have been the energy given to the ball by the throw.

Using

$P=\frac{W}{\Delta t}$

$P=\frac{1.715}{.5}$

A man with mass 50kg is wearing a backpack of mass 5kg and running up a hill with a slope of 30 degrees at a constant velocity of $2\frac{m}{s}$ . What is the man's power output?

$g=10\frac{m}{s^2}$

Explanation

Since the man's kinetic energy remains constant, we are really solving for the rate at which he gains potential energy.

However, when we find the change of this with respect to time (aka power), the expression becomes:

$P = \frac{\Delta U}{t}=\frac{mg\Delta h}{t}$

We can calculate the change in height by using the velocity of the man and the slope of the hill:

$\frac{\Delta h}{t} = v_y = v\cdot sin(30^{\circ})$

Substituting this into the equation for power, we get:

$P = m\cdot g \cdot v\cdot sin(30^{\circ})$

We know all of these values, so we can plug in and solve:

$P = (55)(10)(2)sin(30^{\circ}) = 550 W$

During time period , a rocket ship deep in space of mass $2.55*10^5\textup{ kg}$ travels from $<0,10>\textup{ km}$ to $<5,5>\textup{ km}$ . During time period , the rocket fires. During time period , the rocket travels from $<-15,0>\textup{ km}$ to $<-30,-5>\textup{ km}$ .

Time periods , , and took $10\textup{ s}$ each.

Determine the average power during time period .

$2.55*10^{10}\textup{ W}$

None of these

$1.89*10^{10}\textup{ W}$

$7.14*10^{10}\textup{ W}$

$3.11*10^{10}\textup{ W}$

Explanation

Using

Determining initial kinetic energy:

$v=\frac{d}{t}$

$d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$

Combining equations

$KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$

Converting to and plugging in values:

$KE=.5*2.55*10^5(\frac{\sqrt((5000-0)^2+(5000-10000)^2)}{10})^2$

$KE_i=6.375*10^{10}J$

Determining final kinetic energy:

$v=\frac{d}{t}$

$d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$

Combining equations

$KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$

Converting to and plugging in values:

$KE=.5*2.55*10^5(\frac{\sqrt((-15000+30000)^2+(0+5000)^2)}{10})^2$

$KE_f=3.188*10^{11}J$

Plugging in values:

$W=3.188*10^{11}-6.375*10^{10}$

$W=2.55*10^{11}J$

Using

$J=\frac{W}{t}$

$J=\frac{2.55*10^{11}}{10}$

$J=2.55*10^{10}W$

A car of mass goes from to $100\frac{km}{hr}$ in . Estimate the power of the engine.

Explanation

Power is defined as:

$Power=\frac{Work}{time}$

Calculate work, with only kinetic energy involved:

Convert $\frac{km}{hr}$ to $\frac{m}{s}$ :

$\frac{100km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=27.8\frac{km}{hr}$

$\frac{0km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=0\frac{km}{hr}$

Plug in values:

Plugging in values:

$Power=\frac{5.4*10^5}{7s}$

If the transfer of of energy per second to a object results in the object experiencing a force of and a displacement of , how long will this process take? Assume there is no friction.

Explanation

In this question, we're given a few different variables. We're told there is energy, time, displacement, force, and mass involved. So, we need to think about a way to relate these all together, given what the question is asking.

We're told that energy is being transferred to an object at a certain rate. This energy transfer results in a force on the object which, in turn, causes it to undergo displacement. Finally, we know the mass of the object as well.

Since we're told that the energy is being transferred at a certain rate, we know power is involved. We also know that the change in energy of an object is equal to the force that object experience multiplied by its distance. Thus, we can relate these variables with the following expression.

$P=\frac{W}{t}=\frac{Fd}{t}$

Using this expression, we can rearrange it in order to isolate the term for time.

$t=\frac{Fd}{P}$

Now, we just need to plug the values given in the question stem to solve for our answer.

$t=\frac{(5:N)(12:m)}{20:\frac{J}{s}}=3:sec$

Also notice that we didn't need to use the object's mass to solve this problem. That was extraneous information.

A car goes from $0\frac{m}{s}$ to $45\frac{m}{s}$ in . Estimate how long it would take the car to accelerate to $60\frac{m}{s}$ , assuming the same engine power.

Explanation

Definition of power:

$J=\frac{W}{\Delta t}$

Definition of work:

Since all energy in this case is kinetic:

Initial velocity is zero:

Combine equations:

$J=\frac{.5mv_f^2}{\Delta t}$

Plugg in values for both final velocities and solve for the change in time:

$J=\frac{.5*1500*45^2}{4.5}=\frac{.5*1500*60^2}{\Delta t}$

$\Delta t = 8s$

A locomotive accelerates a train from $0\frac{m}{s}$ to $10\frac{m}{s}$ in . Estimate the power of the locomotive during this time period.

None of these

Explanation

$J=\frac{W}{\Delta t}$

All energy is kinetic, and the initial energy is zero, so:

Combining equations:

$J=\frac{.5mv_f^2}{\Delta t}$

Converting minutes to seconds and plugging in values:

$J=\frac{.5*100*10^4*10^2}{120}$

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure).

This system is initially held so that the rod is in a horizontal position. At , the system is released and the rod is allowed to rotate. How much work is done by gravity by the time mass A reaches the lowest point of rotation?

$g = 10\frac{m}{s^2}$

Neglect air resistance. Assume the mass of the rod is zero.

Explanation

We can use the work-energy theorem to solve this problem

$\Delta K = W$

Now let's use the equation for conservation of energy:

Rearranging for change in kinetic energy, we get:

$K_f - K_i = U_i - U_f = \Delta K$

Substituting this into the work-energy theorem, we get:

$\Delta K= W = U_i - U_f$

We no longer care about the velocity of either mass. We're simply concerned about the change in potential energy.

Using the following equation:

We can expand the later two terms:

$U_i = m_Agh_{A_{i}} + m_Bgh_{B_{i}}$

$U_f = m_Agh_{A_{f}} + m_Bgh_{B_{f}}$

If we assume that the bottom of the circle has a height of 0, we can say that the final potential energy of mass A = 0. Therefore, we get:

$U_i = m_Agh_{A_{i}} + m_Bgh_{B_{i}}$

$U_f = m_Bgh_{B_{f}}$

Substituting these into the work-energy-theorem, we get:

$W = m_Agh_{A_{i}}+m_Bgh_{B_{i}}-m_Bgh_{B_{f}}$

Factoring out g to make things neater:

$W = g\left ( m_Ah_{A_{i}}+m_Bh_{B_{i}}-m_Bh_{B_{f}}\right )$

With our last assumption, we can say that the initial height of both masses

$h_i = \frac{1}{2}l = 1m$

and the final height of mass B

$h_{B_{f}} = 2m$

Now that we have values for everything, it's time to plug and chug:

$W = \left (10\frac{m}{s^2} \right )\left ( 10kg\cdot 1m + 6kg\cdot 1m - 6kg\cdot 2m\right )$

$W = \left ( 10\frac{m}{s^2}\right )\left ( 4kg\cdot m\right )$

You are on the ground floor of a parking garage building. You then run up the stairs to the roof, taking to reach a height of . You have a mass of . Determine your power in watts.