## Mastering Physics Solutions: Pulling a Block on an Incline with Friction

Pulling a Block on an Incline with Friction

Part A = -24.5J
Part B = 700J
Part C = -24.5J
Part D = 700J

Solutions Below:

A block of mass 1.00 kg sits on an inclined plane as shown. A force of magnitude 70.0N is pulling the block up the incline. The coefficient of kinetic friction between the plane and the block is 0.500. The inclined plane makes an angle 60.0 degrees with the horizontal.

Part A

What is the total work Wfric done on the block by the force of friction as the block moves a distance 10.0 m up the incline?

Frictional force is the coefficient of friction times normal force:

FF = μFN

Find normal force first:

FN = mgsin(θ)
FN = 1.00(9.8)cos(60)
FN = 4.9N

Now solve for frictional force:

FF = μFN
FF = 0.500(4.9)
FF = 2.45N

Now solve for work using the formula W = fd:

Wfric = fd
Wfric = 2.45(10)
Wfric = 24.5J

Finally, since friction opposes the direction of motion, we know that the sign must be negative:

Wfric -24.5J

Part B

What is the total work WF done on the block by the applied force F as the block moves a distance 10.0 m up the incline?

Work equals force times distance:

W = Fd
WF = 70.0 * 10.0

WF = 700J

Part C

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

What is the total work WFric done on the block by the force of friction as the block moves a distance 10.0 m down the incline?

The answer is the same as for part A, since the coefficient of friction, normal force, and distance have not changed. Even though the direction changes (from right to left), the sign of the work due to friction is still negative because friction always opposes the direction of motion:

WFric = -24.5J

Part D

What is the total work WF done on the box by the applied force 70.0 N in this case?

Again, since only the direction has changed, the answer remains the same:

WF = 700J

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