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Mathematics

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Question 7

2.5 Points

In the figure, a crate is on a rough surface inclined at . A constant external force is applied horizontally to the crate. While this force pushes the crate a distance of up the incline, its velocity changes from to . How much work does friction do during this process?

(B) +37001

(C) -72001

(D)

(E) zero

Question 7

2.5 Points

In the figure, a

(B) +37001

(C) -72001

(D)

(E) zero

See Answers

Best Answer

The answer is © -72001 J.

We can first use the work-energy principle to find the net work done on the crate:

where is the net work done on the crate, and is the change in its kinetic energy.

Since an external force is applied horizontally, the only force with a component along the direction of motion up the incline is friction, which acts opposite to the direction of motion. Thus, we have:

where is the work done by the external force , is the work done by friction, and is the change in kinetic energy.

The work done by the external force is:

where is the magnitude of the external force, is the distance moved by the crate, and is the angle between the force and the displacement. Since the force is applied horizontally, , so:

J

The change in kinetic energy of the crate is:

where is the mass of the crate, is the initial velocity, and is the final velocity. Substituting the given values, we get:

J

Thus, the net work done on the crate is:

Solving for , we get:

J

Since friction does negative work, the work done by friction is J, which is approximately J. However, the answer choices are in joules, so the correct answer is © -72001 J.

We can first use the work-energy principle to find the net work done on the crate:

where

Since an external force

where

The work done by the external force

where

The change in kinetic energy of the crate is:

where

Thus, the net work done on the crate is:

Solving for

Since friction does negative work, the work done by friction is