A block of weight m g sits on an inclined plane as shown in (Figure 1) . A force

Question

A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is .

Part A

What is the total work Wfric done on the block by the force of friction as the block moves a distance L up the incline?

Express the work done by friction in terms of any or all of the variables , m, g, , L, and F1.

mgLcos

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Part B

What is the total work WF1 done on the block by the applied force F 1 as the block moves a distance L up the incline?

Express your answer in terms of any or all of the variables , m, g, , L, and F1.

F1L

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Now the applied force is changed to F 2, so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed as shown in (Figure 2) .

Part C

What is the total work Wfric done on the block by the force of friction as the block moves a distance L down the incline?

Express your answer in terms of any or all of the variables , m, g, , L, and F2.

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Part D

What is the total work WF2 done on the box by the appled force in this case?

Express your answer in terms of any or all of the variables , m, g, , L, and F2.

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A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is .

Part A

What is the total work Wfric done on the block by the force of friction as the block moves a distance L up the incline?

Express the work done by friction in terms of any or all of the variables , m, g, , L, and F1.

Wfric =

mgLcos

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Correct

Part B

What is the total work WF1 done on the block by the applied force F 1 as the block moves a distance L up the incline?

Express your answer in terms of any or all of the variables , m, g, , L, and F1.

WF1 =

F1L

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Correct

Now the applied force is changed to F 2, so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed as shown in (Figure 2) .

Part C

What is the total work Wfric done on the block by the force of friction as the block moves a distance L down the incline?

Express your answer in terms of any or all of the variables , m, g, , L, and F2.

Wfric =

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Part D

What is the total work WF2 done on the box by the appled force in this case?

Express your answer in terms of any or all of the variables , m, g, , L, and F2.

WF2 =

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L

Explanation / Answer

Coefficiwnt of Kinetic Friceion = mu
Block moves at a constant speed
so, friction actig on the block = mu*N ( where N is the normal force)
from newtons force balance
N = mg*cos(theta)
and f + mg*sin(theta) = F
F = mu*mg*cos(theta) + mg*sin(theta)
F = mg[mu*cos(theta) + sin(theta)] -- (1)

1. WOrk done by friction = -f*L = -mu*mg*cos(theta)*L
2. Work DOne by the force F = F1.L
3. when the force is changed to F2, the block comes down at a constant speed
Work done by friction = -mu*mg*cos(thet)*L
4. Work done by the force F2 = F2.L = F2*L

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