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

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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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Question

A person is moving boxes up the stairs in their new home. They have two identical boxes, with same the size and mass. The first box is easy to carry up the stairs. When moving the second box, the person is more tired and moves more slowly. Which statement accurately describes the work and power between the two trials?

Answer

Because the boxes are the same mass and are moving the same distance, the work done will remain the same between the two instances. Work does not depend on time:

However, when carrying the second box, the person moves more slowly. The overall time increases, which leads to a decrease in power.

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

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Question

When riding your bicycle, you have a maximum power output of 500W. As you approach a hill, you shift into first gear, applying a torque of $30\ N\cdot m$ to the gears. Assuming you produce 50% of your maximum power output, what is the angular velocity of the gears, in radians per second?

Answer

Power is determined by calculating the work output per unit time. In this case, power will be the product of torque and angular velocity:

$P=\tau*\omega$

We are given values for our torque and our power output, allowing us to solve for the angualr velocity. Keep in mind that the power output is going to be 50% of the maximum.

$250W=(30N\cdot m)(\omega)$

$\omega=8.33\frac{rad}{s}$

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Question

A crane lifts a crate with a mass of 50kg. The crate is raised at a constant velocity for ten seconds and and moves a vertical distance upwards of 20m. What power is being supplied to the crane during this time?

Answer

For this problem we can calculate the power as the product of force and velocity:

First, we need to find the velocity:

$v=\frac{\Delta x}{\Delta t}=\frac{20m}{10s}=2\frac{m}{s}$

Our force will be equal to the weight of the crate:

$F=mg=(50kg)(9.8\frac{m}{s})=490N$

Now, we can solve for power:

$P=Fv=(490N)(2\frac{m}{s})=980W$

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Question

A 25kg child climbs up a tree. How much work is required for him to climb up this tree to a height of three meters?

Answer

The forces acting on the child are the force of gravity and the upward force provided by the child himself. Choosing the upward direction as positive, Newton's second law applied to the child gives the following equation.

$\Sigma F_y=F_c-mg=0$

$F_c=mg=(25\ \text{kg})(9.8\ \frac{m}{s^2})=245\ \text{N}$

To calculate the work done by the child to bring himself three meters up the tree, we use the work equation below.

$W=F_c\cdot d=(245\ \text{N})(3\ \text{m})=735\ \text{J}$

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Question

A force of 27N is applied horizontally to a box resting on a level surface. The surface exerts a friction force of 2N as the box moves. If it moves a total of 3.7m, how much work was done on the box?

Answer

The net force on the box is the applied force minus the friction force, since friction acts in the direction opposite motion.

$F_{net}=F_{app}-F_{friction}$

Consequently, the net force is:

$F_{net}=27N-2N = 25 N$

The formula for work is:

$W=F_{net}*d$

Substituting in our net force and distance, we find that the work done on the box is:

$W=25N(3.7m)=92.5 N\cdot m$

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Question

An object is pushed across a rough surface with a force of 53N. The rough surface exerts a frictional force of 3.47N on the object. If the object is pushed 7.9m, how much work is done on the object?

Answer

The defintion of work is:

$W=F_{net}\cdot d$

The net force on this object is:

$F_{net}=F_{app}-F_{friction}$

We can calculate this term using the given values:

$F_{net}=53N-3.47N=49.53N$

The distance is given. Substituting these values:

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Question

ONe of the forces that act on a particle as it underjoes a dispalcement of

m is given by .

How much work is done by that force on the particle?

Answer

The equation of work is given by $W = F\cdot d$ , where is force and is the distance. Both the work and the force are given to us, but in vector form. In this case, we have to take the dot product of the force and distance.

When taking dot product, keep these rule in mind

$i \cdot i = 1, j \cdot j = 1, k \cdot k =1$

Every other dot proudct is equal to 0.

$(8i - 2j +3k) \cdot (2i + 3j - 1k) = \((8\cdot2)+(-2\cdot3)+(3\cdot-1) ) = 7:J$

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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.

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

$v_f = \sqrt{900}$

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

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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.

$v = \sqrt{2gh}$

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

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

Compare your answer with the correct one above

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.

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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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