Pulling a Block on an Incline with Friction A block of weight m g sits on an inc

Question

Pulling a Block on an Incline with Friction

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 Lup the incline?

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

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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 Lup the incline?

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

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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 Ldown 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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Pulling a Block on an Incline with Friction

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 Lup the incline?

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

Wfric =

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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 Lup the incline?

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

WF1 =

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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 Ldown 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

Given

mass of box is m, as the box is applied by a force F , up the incline

then the forces acting horizontally are

downward the frictional force and the force of gravity component, the applied force is upward the incline

so writing the force equation taking downward is -ve and upward is +ve

   -(Ff + Fg) +F1 = 0

   Ff+Fg = F1

   mg sin theta + muek mg cos theta = F1

PArtA

then the workdone by the friction force about the distance L is

   Wfric = mue_k. mg cos theta*L

Part B

   total work done by the applied force is

   WF = F1*L cos theta
   WF1 = (mg sin theta + muek mg cos theta)(Lcos theta)

PArt C

now the box is pushed downward instead of pull in the upward direction then

then the forces acting horizontally are

downward the applied force and the force of gravity component, the frictional force is upward the incline

so writing the force equation taking downward is -ve and upward is +ve

   -(F2 + Fg) +Ff = 0

   F2+Fg = ff

   F2 + mg sin theta = muek mg cos theta

Part C

   the workdone by frictional force is Wf = muek*mg cos theta*L

Part D

   work done by the appllied force is WF2 = F2*L cos theta

               WF2 = ((muek mg cos theta)-(mg sin theta))(L)cos theta

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