Work, Energy, and Power

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AP Physics 1 › Work, Energy, and Power

Questions 1 - 10

A car of mass is accelerated from $10 \frac{m}{s}$ to $35 \frac{m}{s}$ in 2s.

Determine the work done on the car in this time frame.

None of these

Explanation

Use the definition of work:

$W=\Delta KE$

Plug in known values and solve.

Terry believes he can throw a ball vertically and hit a target above himself. If the ball weighs , how fast must it be traveling when it leaves his hand to just reach the target? Neglect air resistance.

$g=9.81\frac{m}{s^2}$

$31.32\frac{m}{s}$

$15.66\frac{m}{s}$

$22.15\frac{m}{s}$

$981\frac{m}{s}$

$245.25\frac{m}{s}$

Explanation

This problem deals with both potential energy and kinetic energy.

Potential energy is expressed as:

Kinetic energy is expressed as:

$KE=\frac{1}{2}mv^2$

Energy must be conserved, so set up the following equation:

The initial height can be treated as zero, as can the final velocity. Plug in these zero values into the above equation.

Solve for , the initial velocity the ball needs in order to reach a maximum height of .

$v_1=\sqrt{2gh_2}$

$v_1=\sqrt{2\left(9.81\frac{m}{s^2}\right)(50m)}$

$v_1=31.32\frac{m}{s}$

How much work (in kilojoules) is done to accelerate a car (3000kg) from rest to $60\frac{m}{s}$ .

5400 kilojoules

3000 kilojoules

6200 kilojoules

3400 kilojoules

2300 kilojoules

Explanation

Work is found by finding the change in kinetic energy. Since the car started from rest it had no initial kinetic energy.

$KE=\Delta.5(3000kg)(60)^2=5400000J$

Divide by 1000 to convert to kilojoules and get 5400 kilojoules.

A car is at rest at the bottom of a hill. later, it is at the top of the hill going $100\frac{km}{hr}$ . Find the net work done.

Explanation

Initially the car is at rest at the bottom of a hill, this velocity and height are zero.

Converting $\frac{km}{hr}$ to $\frac{m}{s}$

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

Plugging in values:

An 500kg elevator is at rest. If it is raised 50 meters and returns to rest, how much total work was done on the elevator?

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

Explanation

This can be a tricky question. You need to rely on the work-energy theorem, which states:

$W_{net}= \Delta K$

Since the elevator is at rest at both the beginning and end, the net work is 0; there is no net change in energy, and therefore no work.

This theorem can be confusing to some since it completely negates potential energy. However, let's think about the situation presented in the problem. A force is required to raise the elevator, meaning that energy is put into the system. However, since it comes back to rest, all of the energy that was put in has been removed by the force of gravity, resulting in a net of zero work.

As a joke, Charlie glues C.J's phone to its receiver, which is bolted to her desk. Trying to extricate it, C.J. pulls on the phone with a force of for . She then pulls on the phone with a force of for . Unfortunately, all of her exertion is in vain, and neither the phone, nor receiver move at all. How much work did C.J. do on the phone in her 25 total seconds of pulling?

Explanation

Work is a measure of force and displacement $(W = F \cdot s)$ . Because C.J. did not move the phone at all, no work was done.

A rocket is in space at location when it fires it's thrusters. The thrusters provide a force of . The thusters are turned off at location .

What is the work done on the rocket by the thrusters?

$5.1\times10^5 J$

None of these

Explanation

Because all of the force was in the Y direction, all of the work will come from the change in the Y coordinate.

We will use the definition of work

$W=F\cdot d$

First we need to find the distance traveled.

Then we can plug that into our work equation.

$52 m\cdot 9900 N = 5.1\times10^5 J$

A semi-truck carrying a trailer has a total mass of . If it is traveling up a slope of $5^{\circ}$ to the horizontal at a constant rate of $20\frac{m}{s}$ , how much power is the truck exerting?

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

Explanation

Since the truck is traveling at a constant rate, we know that all of the power exerted by the truck is going into a gain in potential energy. The power exerted will be a function of the change in potential energy over time. Therefore, we can write the following formula:

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

is a vertical height, so we need to write that as a function of distance traveled up the slope:

$h = d\cdot sin (\theta)$

$P = \frac{mgd\cdot sin(\theta)}{t}$

We can substitute velocity into this equation:

$P = mgv\cdot sin(\theta)$

We have values for all of these variables, allowing us to solve:

$P = (1500kg)(10\frac{m}{s^2})(20\frac{m}{s})\left sin(5^{\circ})\right = 26000W$

Determine the work done by nonconservative forces if an object with mass 10kg is shot up in the air at $30\frac{m}{s}$ returns to the same height with speed $27\frac{m}{s}$ .

Explanation

Since this question refers to work done by nonconservative forces, we know that:

$W_{outside}=\Delta PE+\Delta KE$

Here, $\Delta PE$ is the change in potential energy, and $\Delta KE$ is a change in kinetic energy. $\Delta PE$ is because the object returns to the same height as when it was launched. $\Delta KE$ however has changed because the object's velocity has changed. Recall that the formula for the change in kinetic energy is given by:

$\Delta KE= mv^2_{final}-mv^2_{initial}$

Here is the mass of the object, $v_{final}$ is the final velocity of the object and $v_{initial}$ is the initial velocity of the object.

In our case:

, $v_{final}=45 \frac{m}{s}$ , and $v_{initial}=50 \frac{m}{s}$

$\Delta KE= mv^2_{final}-mv^2_{initial}$

$\Delta KE=10kg*\left(27\frac{m}{s}\right)^2-10kg*\left(30\frac{m}{s}\right)^2$

$\Delta KE= 7290J-9000J= -1710J$

$W_{outside}=\Delta PE+\Delta KE$

$W_{outside}=-1710J$

The work done by a centripetal force on an object moving in a circle at constant speed is __________.

zero

equal to the kinetic energy of the object

equal to the force exerted

equal to the force exerted multiplied by the displacement

Explanation

Recall that work can be defined as:

$W=F*d*cos(\theta)$

Here, is the magnitude of the force vector, is the magnitude of the displacement vector, and $\theta$ is the angle between the directions of the force and displacement vectors. In the case of circular motion, the force vector is normal to the circle since it points inward, and the displacement vector is tangent to the circle. This means that the angle between the force vector and displacement is . Since , work done by the centripetal force on an object moving in a circle is always .

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