A block of mass 1.00 sits on an inclined plane as shown. A force of magnitude 80

Question

A block of mass 1.00 sits on an inclined plane as shown. A force of magnitude 80.0 is pulling the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is 0.500. The inclined plane makes an angle 10.0 with the horizontal.

A) What is the total work Wf done on the block by the applied force F vector as the block moves a distance 9.00m up the incline?

Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed.

B) What is the total work Wf done on the block by the applied force F as the block moves a distance 9.00m up the incline?

C) What is the total work Wfric done on the block by the force of friction as the block moves a distance 9.00m down the incline?

D)What is the total work WF done on the box by the applied force 80.0N in this case?

Explanation / Answer

A) Given F = 80 N

d = 9 m

Since force is in the direction of displacement

So work (W) = 80*9 = 720 J

B) Here since the direction of force and displacement is opposite.

So, W = -F*d = -80 * 9 = -720 J

C) Friction = Friciton coefficient * Normal force

Friction coefficient = 0.5

Normal force = Mg Cos(10) = 1 * 9.8 COs(10) = 9.65 N

Friction force (fr) = 9.65 N

As the block moves down the incline, direction of frictional force is opposite to displacement. So,

W = - fr * d = -9.65 * 9 = 86.86 J

D) Since the direction of force and displacement is same

W = F*d = 80 * 9 = 720 J

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