Work, Energy, and Power - AP Physics C: Electricity and Magnetism

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Question

A train car with a mass of 2400 kg starts from rest at the top of a 150 meter-high hill. What will its velocity be when it reaches the bottom of the hill, assuming that the bottom of the hill is the reference level.

Answer

The law of conservation of energy states:

If the car starts at rest, then the initial kinetic energy = 0 J.

If the car ends at the reference height, the final potential energy = 0 J.

Subsituting these values, the equation becomes:

The initial potential energy can be determined by:

$PE_i = mgh_i \ PE_i = 2400 kg \cdot 9.81 \frac{m}{s^2} \cdot 150 m\ PE_i = 3531600 \frac{kg \cdot m^2}{s^2}\ PE_i = 3531600 J$

The final kinetic energy equation is:

$KE_f = \frac{1}{2}mv_f^2\ KE_f = \frac{1}{2}(2400 kg)v_f^2$

Substituting the initial potential energy and final kinetic energy into our modified conservation of energy equation, we get:

$3531600 J = \frac{1}{2}(2400 kg)v_f^2\ 3531600 \frac{kg\cdot m^2}{s^2} = 1200 kg \cdot v_f^2\ \frac{3531600 \frac{kg\cdot m^2}{s^2}}{1200 kg}=v_f^2\ 2943 \frac{m^2}{s^2}=v_f^2\ \sqrt{2943 \frac{m^2}{s^2}}=\sqrt{v_f^2}\ 54.25 \frac{m}{s}=v_f$

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Question

A 120kg box has a kinetic energy of 2300J. What is its velocity?

Answer

The formula for kinetic energy of an object is:

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

The problem gives us the mass and the kinetic energy, and asks for the velocity, so we can rearrange the equation:

$v=\sqrt{\frac{2KE}{m}}$

Use our given values for kinetic energy and mass to solve:

$v=\sqrt{\frac{2(2300J)}{120kg}}=6.2\frac{m}{s}$

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Question

An object has a mass of 5kg and has a position described by the given function:

What is the object's kinetic energy after two seconds?

Answer

Kinetic energy is defined by the equation:

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

Taking the derivative of the position function allows us to obtain the velocity function:

We can now determine the velocity after two seconds:

Now that we know our velocity, we can solve for the kinetic energy.

$KE=\frac{1}{2}(5kg)(3\frac{m}{s})^2=22.5J$

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Question

An object starts from rest and accelerates at a rate of $2.5\frac{m}{s^2}$ . If the object has a mass of 10kg, what is its kinetic energy after three seconds?

Answer

Kinetic energy is given by the equation:

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

We can find the velocity using the given acceleration and time:

$v=v_o+a\Delta t$

$v=0\frac{m}{s}+(2.5\frac{m}{s^2})(3s)=7.5\frac{m}{s}$

Use this velocity to find the kinetic energy after three seconds:

$KE=\frac{1}{2}(10kg)(7.5\frac{m}{s})^2=281.25J$

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Question

Calculate how much potential energy a vertical standing spring gains if someone puts a 4kg box on top of it. Take the spring constant to be 120N/s.

Answer

First, find out how much the spring compresses when the 4kg box is put on top of it. To find this, consider what forces are acting on the box when it is resting on the spring. The forces acting on it are the upward spring force, , and the downward gravitational force,; thus, the net force acting on the box is given by the equation below.

$F_{net}=F_s-F_g=-kx-mg=0$

We have chosen the upward direction to be positive and the downward direction to be negative. The net force is equal to zero because the box is at rest. Solve for x.

$x=-\frac{mg}{k}$

$x=-\frac{(4\ kg)(-9.8\ \frac{m}{s^2})}{120\ \frac{N}{m}}$

$x=0.33\ \text{m}$

Then use this value to find the potential energy of the spring.

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}(120\ \frac{N}{m})(0.33\ m)^2$

$U=6.5\ J$

So when the box is placed on top of the spring, the spring gains a potential energy of 6.5J.

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Question

A man throws a 5kg ball straight up. At the top of its trajectory, the ball has a potential energy of 160 Joules. At what velocity does the man initially throw the ball?

Answer

We can use conservation of energy to find the velocity required to launch the ball.

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

$PE = KE \rightarrow 160 = \frac{1}{2}mv^2$

$v = 8\ \frac{m}{s}$

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Question

A man throws a 5kg ball straight up. At the top of its trajectory, the ball has a potential energy of 160 Joules. What is the highest point the ball reaches if it is thrown from the level of the ground?

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

Answer

The equation for potential energy is:

We can rearrange to solve for the height:

$h=\frac{PE}{mg}$

Plug in our given values to solve:

$h=\frac{160J}{(5kg)(10\frac{m}{s^2})}=3.2m$

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Question

$G \approx (20/3) 10^{-11} \ \frac{m^3}{kgs^2}, \quad M_{earth} \approx 610^{24}\ kg, \quad R_{earth} \approx 6000 \ km$

A man launches a 5kg ball straight up from sea level. Disregarding drag, if the man would like to launch the ball into an escape trajectory from earth's gravity, approximately how fast must he launch the ball (in meters per second)?

$G=6.6710^{-11}\frac{m^3}{s^2kg}$

$M_E=6*10^{24}kg$

Answer

In order to launch the ball at a velocity that escapes Earth's gravity, the kinetic energy must overcome the gravitational pull of the Earth. We can use conservation of energy to make the kinetic energy and gravitational potential energies equivalent.

$KE_{escape} = \frac{1}{2}m v_{escape}^2$

$U_{gravitational} = \frac{GMm}{r}$

$KE_{e} = U_{g}$

$\frac{1}{2}mv_{e}^2= \frac{GMm}{r}$

Rearrange to solve for the escape velocity:

$v_e = \sqrt{\frac{2GM}{r}}$

Finally, plug in our values and solve.

$v_e= \sqrt{\frac{2(6.6710^{-11})(610^{24})}{610^6}}$

$v_e=\sqrt{\frac{72}{6}10^{7}} =\sqrt{1.210^{8}} \approx\sqrt{12110^{6}} = 11000\ \frac{m}{s}$

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Question

A 0.8kg ball is dropped from rest from a cliff that is 150m high. Use conservation of energy to find the vertical velocity of the ball right before it hits the bottom of the cliff.

Answer

The conservation of energy equation is .

The ball starts from rest so . It starts at a height of 150m, so . When the ball reaches the bottom, height is zero and thus, and $K_2=\frac{1}{2}mv^2$ . The conservation of energy equation can be adjusted below.

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

Solve for v.

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8\ \frac{m}{s^2})(150\ m)}$

$v=54.2\ \frac{m}{s}$

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Question

Starting from rest, a skateboarder travels down a 25o incline that's 22m long. Using conservation of energy, calculate the skateboarder's speed when he reaches the bottom. Ignore friction.

Answer

Conservation of energy states that .

The skateboarder starts from rest; thus, and . At the bottom of the incline, $K_2=\frac{1}{2}mv^2$ and .

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

Solve for v.

$v=\sqrt{2gh}$

Using trigonometry, $h=22\sin(25)$ .

$v=\sqrt{2(9.8\ \frac{m}{s^2})(22)(\sin(25))}$

$v=13.5\ \frac{m}{s}$

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Question

A $\small 5.0kg$ bowling ball is dropped from $\small 5.0m$ above the ground. What will its velocity be when it is $\small 1.0m$ above the ground?

Answer

Relevant equations:

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

Determine initial kinetic and potential energies when the ball is dropped.

$U_i = (5kg)(9.8\frac{m}{s^2})(5m)=245J$

Determine final kinetic and potential energies, when the ball has fallen to $\small 1.0m$ above the ground.

$K_f=\frac{1}{2}(5kg)v_f^2$

$U_f = (5kg)(9.8\frac{m}{s^2})(1m)=49J$

Use conservation of energy to equate initial and final energy sums.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

Solve for the final velocity.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

$v_f^2=\frac{245J-49J}{\frac{1}{2}(5kg)}=78.4\frac{m^2}{s^2}$

$v_f=8.85 \frac{m}{s} \approx 8.9\frac{m}{s}$

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Question

A solid metal object with mass of is dropped from rest at the surface of a lake that is deep. The water exerts a drag force on the object as it sinks. If the total work done by the drag force is -, what is the speed of the object when it hits the sand at the bottom of the lake?

Answer

This is a conservation of energy problem. First we have to find the work done by gravity. This can be found using:

$W= F\cdot d, F = mg \ W = 25:kg \cdot 9.8:m/s^2 \cdot 20:m = 4900:J$

It is given to us that the work done by the drag force is which means that work is done in the opposite direction. We take the net work by adding the two works together we get, of net work done on the block.

Since this is a conservation of energy problem, we set the net work equal to the kinetic energy equation:

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

is the mass of the block and we are trying to solve for .

$v = \sqrt{\frac{2 \cdot 1100}{25}} = 9.38:m/s$

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Question

If a roller coaster car is traveling at when it is above the ground, how fast is it going when above the ground?

Answer

This is a classic conservation of energy problem. We know that potential energy and kinetic energy both have to conserve. So we use the following equation:

$\frac{1}{2}mv_o^2+mgh_1 = \frac{1}{2}mv_f^2 + mgh_2$

What this equations says is, the sum of kinetic and potential energy is same at varying heights and velocities.

We can simplify this equation by cancelling out all the m terms.

$\frac{1}{2}v_o^2+gh_1 = \frac{1}{2}v_f^2 + gh_2$

We know all the terms except for , which is the final speed we are trying to solve for , which is , is and is .

If we plug in all the numbers and solve for , we get .

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Question

An object with a mass of is moving at $25\frac{m}{s}$ in a straight line on a fricitonless surface. After a force of acting in the direction of its motion is applied to it for , what is the object's speed in meters per second?

Answer

Begin by using the following equation relating the initial and final kinetic energy and the work done on the object:

Then, plug in the given variables and solve for the final speed.

$\frac{1}{2}mv_i^2 + Fd = \frac{1}{2}mv_f^2$

$\frac{1}{2}(25kg)(25\frac{m}{s})^2 + (250N)(13.75m) = \frac{1}{2}(25kg)v_f^2$

Simplify terms.

$7812.5 + 3437.5 = \frac{25}{2}v_f^2$

Isolate the final velocity and solve.

$v_f = \sqrt{11250*\frac{2}{25}}$

$v_f = \sqrt{900}$

$v_f = 30\frac{m}{s}$

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Question

A projectile is launched straight upwards at an initial velocity of $32.65\frac{m}{s}$ . What is the maximum height that this projectile reaches in meters?

Round to the nearest meter, and assume the projectile encounters no air resistance.

Answer

You can use the motion equation and find the maximum, but it may be faster to use energy equations. Set the initial kinetic energy equal to the gravitational potential energy at the maximum height and solve for the height.

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

Mass cancels.

$\frac{1}{2}v^2 = gh$

Isolate the height and solve.

$h = \frac{v^2}{2g} = \frac{(32.65\frac{m}{s})^2}{2(9.8\frac{m}{s^2})} = 54.3889m$

Round to .

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Question

A set of cars on a roller coaster with a combined mass of is at the top of its initial hill and will drop down the hill before the track starts to rise again. What will the coaster's speed be at the moment the track starts to rise again?

Round to the nearest meter per second. You may also assume the track does not create friction.

Answer

Remember that gravitational potential energy is not affected by the path downward (or upward)—whether it is straight, curved, or winding—only by how big the drop is. Once that is taken into account, you can simply set the initial gravitational potential energy and final kinetic energy equal to each other as if the coaster were falling straight down and solve for the final velocity.

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

The mass cancels.

$gh = \frac{1}{2}v^2$

Isolate the velocity and solve.

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8\frac{m}{s^2})(45m)}$

$v = 29.7\frac{m}{s}\approx30\frac{m}{s}$

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Question

What power is required to lift a 25kg box 5.76m off the ground in fifteen seconds?

Answer

The definition of power is:

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

The work done on the box to lift it is required in order to overcome the force of gravity. If we can find the force of gravity on the object, we can calculate the net force. The force of gravity on any object near the Earth's surface is:

The definition of work is:

$W=F_{net}d$

We can substitute the force of gravity for the net force, resulting in the equation:

Substituting this into our power equation, we find:

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

Plugging in our given values and constants, we find:

$P=\frac{(25 kg) (9.81 \frac{m}{s^2})(5.76m)}{15 s}$

$P=\frac{(25)(9.81)(5.76)}{15}\frac{kgm^2}{s^3}$

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Question

A plane weighing 1500kg dives 40m with its engine off before it goes into a circular pattern with a radius of 200m while maintaining its speed at the end of its dive. How much centripetal force, in Newtons, is acting on the plane?

Answer

First, find the gravitational potential energy of the drop. Then, set it equal to the kinetic energy at the end of the drop and solve for the velocity.

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

The mass cancels.

$gh = \frac{1}{2}v^2$

Isolate the velocity and solve.

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8\frac{m}{s^2})(40m)}$

$v = 28\frac{m}{s}$

This gives you the last term you need to solve for the centripetal force.

$F_c = \frac{mv^2}{r}$

$F_c = \frac{(1500kg) (28\frac{m}{s})^2}{200m} = 5880N$

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Question

If the maximum speed of an object attached to the end of a elastic has a 1:1 ratio (a meter per second for each meter) with how much the elastic is stretched or compressed from its starting position, which of the following is true?

Answer

Set the elastic potential and kinetic energy equations equal to each other:

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

You are given the fact that in this case, . This allows you to simplify the equality.

$\frac{1}{2}kx^2 = \frac{1}{2}mx^2$

$\frac{1}{2}k = \frac{1}{2}m$

This shows us that there is a 1:1 ratio between the spring constant of the elastic and the mass of the object.

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Question

A pumpkin is being launched out of an air cannon. For safety reasons, the pumpkin cannot be more than off the ground during flight, and this particular cannon always launches pumpkins at $80\frac{m}{s}$ meters per second—any more power and the pumpkin could be blasted apart; any less and the pumpkin may not leave the launch tube.

What is the maximum possible angle of launch in degrees?

Round to the nearest whole degree.

Answer

We know the maximum height of the pumpkin, which tells us the maximum energy of the launch. Calculate the final gravitational potential energy.

$PE_g = mgh = (10kg ) (9.8\frac{m}{s^2}) (100m) = 9800J$

Now set this value equal to a kinetic energy equation that uses the vertical component of the velocity at launch.

$9800 = \frac{1}{2}mv_y^2$

$v_y = \sqrt{\frac{9800J * 2}{10kg}}$

$v_y = 44.27\frac{m}{s}$

Use the vertical velocity component to determine the launch angle.

$\theta = \sin^{-1}(\frac{v_y}{v})$

$\theta= \sin^{-1}(\frac{44.27\frac{m}{s}}{80\frac{m}{s}}) = 33.60^\circ\approx34^\circ$

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