An applied force of 122 N is exerted horizontally on an 28 kg box of dogs in Yulin. The coefficient of kinetic friction between the floor and the box is 0.35. If the box is initially at rest, what is the final kinetic energy after it has been moved 9.46 m across the floor?

Question

Physics

Roselyn Larson

1 July, 21:28

An applied force of 122 N is exerted horizontally on an 28 kg box of dogs in Yulin. The coefficient of kinetic friction between the floor and the box is 0.35. If the box is initially at rest, what is the final kinetic energy after it has been moved 9.46 m across the floor?

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Answers (1)

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Isabela Watkins · Answer 1

The final kinetic energy is 242 J.

Explanation:

Hi there!

According to the work-energy theorem, the total work done on the box is equal to its change in kinetic energy:

W = ΔKE

Where:

W = work done on the box.

ΔKE = change in kinetic energy (final KE - initial KE).

The only forces that do work in this case are the applied force and the friction force because the box moves only horizontally.

The equation of work is the following:

W = F · s

Where:

F = force.

s = traveled distance.

Then, the work done by the applied force is:

W = 122 N · 9.46 m = 1.15 * 10³ J

To calculate the work done by friction, we have to find the friction force:

Fr = N · μ

Where:

Fr = friction force.

N = normal force.

μ = coefficient of kinetic friction.

The box does not have a net vertical acceleration. It means that the sum of the vertical forces acting on the box is zero:

∑Fy = 0

In this case, the only vertical forces are the weight of the box and the normal force. Then:

Weight + N = 0

N = - Weight

The weight of the box is calculated as follows:

Weight = m · g

Where:

m = mass of the box.

g = acceleration due to gravity.

Then:

-Weight = N = 28 kg · 9.8 m/s² = 274.4 N

Now, we can calculate the friction force:

Fr = N · μ

Fr = 274.4 N · 0.35 = 96 N

The work done by the friction force will be:

W = Fr · s

W = 96 N · 9.46 m = 908 J

Since the work done by friction opposes to the sense of movement, the work is negative.

Now, we can calculate the total work done on the box:

W total = W applied forece + W friction force

W = 1.15 * 10³ J - 908 J = 242 J

Applying the work-energy theorem:

W = final KE - initial KE

Since the box is initially at rest the initial kinetic energy is zero. Then:

W = final KE - 0

W = final KE

Final KE = 242 J

The final kinetic energy is 242 J.