Pulling a Block on an Incline with Friction

Part A = -24.5J

Part B = 700J

Part C = -24.5J

Part D = 700J

Solutions Below:

Part A

_{fric}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:

F_{F} = μF_{N}

Find normal force first:

F_{N} = mgsin(θ)

F_{N} = 1.00(9.8)cos(60)

F_{N} = 4.9N

Now solve for frictional force:

F_{F} = μF_{N}

F_{F} = 0.500(4.9)

F_{F} = 2.45N

Now solve for work using the formula W = fd:

W_{fric} = fd

W_{fric} = 2.45(10)

W_{fric} = 24.5J

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

W_{fric} -24.5J

Part B

_{F}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

W_{F} = 70.0 * 10.0

W_{F} = 700J

Part C

What is the total work W_{Fric} 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:

W_{Fric} = -24.5J

Part D

_{F}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:

W_{F} = 700J