AP Physics C: Mechanics : Work, Power, and Diagrams

Study concepts, example questions & explanations for AP Physics C: Mechanics

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Example Questions

Example Question #1 : Work, Power, And Diagrams

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?

Possible Answers:

Correct answer:

Explanation:

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.

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

Example Question #2 : Work, Power, And Diagrams

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?

Possible Answers:

Correct answer:

Explanation:

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

Consequently, the net force is:

The formula for work is:

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

Example Question #3 : Work, Power, And Diagrams

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?

Possible Answers:

Correct answer:

Explanation:

The defintion of work is:

The net force on this object is:

We can calculate this term using the given values:

The distance is given. Substituting these values:

Example Question #3 : Work, Energy, And Power

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?

Possible Answers:

Correct answer:

Explanation:

The equation of work is given by , 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

Every other dot proudct is equal to 0.

Example Question #4 : Work, Energy, And Power

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  to the gears. Assuming you produce 50% of your maximum power output, what is the angular velocity of the gears, in radians per second? 

Possible Answers:

Correct answer:

Explanation:

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

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.

Example Question #4 : Work, Power, And Diagrams

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? 

Possible Answers:

Correct answer:

Explanation:

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

First, we need to find the velocity:

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

Now, we can solve for power:

Example Question #5 : Work, Power, And Diagrams

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?

Possible Answers:

Work increases; power decreases

Work remains the same; power decreases

Work remains the same; power increases

Work decreases; power stay the same

Work increases; power increases

Correct answer:

Work remains the same; power decreases

Explanation:

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.

Example Question #2 : Power

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

Possible Answers:

Correct answer:

Explanation:

The definition of power is:

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:

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

Substituting this into our power equation, we find:

Plugging in our given values and constants, we find:

Example Question #1 : Interpreting Work, Energy, And Power Diagrams

An object with a mass of  is moving at  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?

Possible Answers:

Correct answer:

Explanation:

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.

Simplify terms.

Isolate the final velocity and solve.

Example Question #1 : Interpreting Work, Energy, And Power Diagrams

 projectile is launched straight upwards at an initial velocity of . What is the maximum height that this projectile reaches in meters?

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

Possible Answers:

Correct answer:

Explanation:

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.

Mass cancels.

Isolate the height and solve.

Round to .

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