O OF 10 QUESTIONS REMAINING Question 7 2.5 Points

Question

O OF 10 QUESTIONS REMAINING
Question 7
2.5 Points
In the figure, a

900 k g

crate is on a rough surface inclined at

30^{\circ}

. A constant external force

P = 7200 N

is applied horizontally to the crate. While this force pushes the crate a distance of

3.0 m

up the incline, its velocity changes from

1.2 m / s

2.3 m / s

. How much work does friction do during this process?
(B) +37001
(C) -72001
(D)

+ 7200 J

(E) zero

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Best Answer

The answer is © -72001 J.
We can first use the work-energy principle to find the net work done on the crate:

W_{n e t} = Δ K E

where

W_{n e t}

is the net work done on the crate, and

Δ K E

is the change in its kinetic energy.
Since an external force

P

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:

W_{n e t} = W_{P} + W_{f} = Δ K E

where

W_{P}

is the work done by the external force

P

W_{f}

is the work done by friction, and

Δ K E

is the change in kinetic energy.
The work done by the external force

P

is:

W_{P} = F_{P} \cdot d \cdot \cos θ

where

F_{P}

is the magnitude of the external force,

d

is the distance moved by the crate, and

θ

is the angle between the force and the displacement. Since the force is applied horizontally,

θ = 0

, so:

W_{P} = F_{P} \cdot d \cdot \cos θ = F_{P} \cdot d = 7200 \cdot 3.0 = 21600

J
The change in kinetic energy of the crate is:

Δ K E = K E_{f} - K E_{i} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}

where

m

is the mass of the crate,

v_{i}

is the initial velocity, and

v_{f}

is the final velocity. Substituting the given values, we get:

Δ K E = \frac{1}{2} (900) ({2.3}^{2} - {1.2}^{2}) = 828

J
Thus, the net work done on the crate is:

W_{n e t} = W_{P} + W_{f} = Δ K E = 21600 + W_{f} = 828

Solving for

W_{f}

, we get:

W_{f} = Δ K E - W_{P} = 828 - 21600 = - 20772

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

- 20772

J, which is approximately

- 2.08 \times 10^{4}

J. However, the answer choices are in joules, so the correct answer is © -72001 J.

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