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

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 Wric 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 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. 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) .

Explanation / Answer

Given,
Weight = m*g
Angle =
Coefficient of Friction = u

PART A
Work done = Force * distance moved
Friction acts allways in the opposite direction , Therefore work done by Friction is Negative.
Work done by Friction, Wfric = - u*mg*cos()*L

PART B
Total Work done = Force * distance moved
Total Work done = (m*g*sin() + u*mg*cos() ) * L
Total Work done, WF1 = m*g*(sin() + u*cos()) * L

PART C
Work done = Force * distance moved
Friction acts allways in the opposite direction , Therefore work done by Friction is Negative.
Work done by Friction, Wfric = - u*mg*cos()*L

PART D
Total Work done = Force * distance moved
Total Work done. WF2 = F2*L
Where, F2 = u*mg*cos() - mg*sin()

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