A block of weight w sits on a plane inclined at an angle θas shown. (Figure 1) The coefficient of kinetic friction between the plane and the block is μ.
Part A
What is the work Wf done on the block by the force of friction as the block moves a distance L up the incline?
Express your answer in terms of some or all of the following: μ, w, θ, L.
Part B
What is the work W done by the applied force of magnitude F?
Express your answer in terms of some or all of the following: μ, w, θ, L.
Part C
What is the change in the potential energy of the block, ΔU, after it has been pushed a distance L up the incline?
Express your answer in terms of some or all of the following: μ, w, θ, L.
The Correct Answer and Explanation is:
To solve the problem, we need to analyze the forces acting on the block and how they contribute to the work done in each part.
Part A: Work Done by the Force of Friction
The frictional force, ( f_k ), opposes the motion of the block as it moves up the incline. The magnitude of the kinetic frictional force is given by:
[
f_k = \mu N
]
where ( \mu ) is the coefficient of kinetic friction and ( N ) is the normal force. On an inclined plane, the normal force is given by:
[
N = w \cos(\theta)
]
where ( w ) is the weight of the block and ( \theta ) is the angle of inclination. Therefore, the frictional force becomes:
[
f_k = \mu w \cos(\theta)
]
Since the frictional force opposes the motion, the work done by the frictional force ( W_f ) is:
[
W_f = – f_k L
]
Substituting the expression for ( f_k ):
[
W_f = – \mu w \cos(\theta) L
]
This is the work done by the frictional force as the block moves up the incline.
Part B: Work Done by the Applied Force
Let ( F ) represent the applied force that moves the block up the incline. The applied force must overcome both the frictional force and the component of the weight of the block that acts parallel to the incline. The parallel component of the gravitational force is:
[
w \sin(\theta)
]
Thus, the total force required to move the block up the incline is:
[
F_{\text{total}} = f_k + w \sin(\theta)
]
Substituting the expression for ( f_k ):
[
F_{\text{total}} = \mu w \cos(\theta) + w \sin(\theta)
]
The work done by the applied force ( W ) is the applied force multiplied by the distance traveled:
[
W = F_{\text{total}} L = (\mu w \cos(\theta) + w \sin(\theta)) L
]
Part C: Change in Potential Energy of the Block
The change in potential energy of the block is given by the difference in gravitational potential energy as the block moves a height ( h ) along the incline. The height gained by the block is:
[
h = L \sin(\theta)
]
The change in potential energy is:
[
\Delta U = w h = w L \sin(\theta)
]
Thus, the change in potential energy of the block is:
[
\Delta U = w L \sin(\theta)
]
Summary of Answers:
- Part A: The work done by the force of friction is:
[
W_f = – \mu w \cos(\theta) L
]
- Part B: The work done by the applied force is:
[
W = (\mu w \cos(\theta) + w \sin(\theta)) L
]
- Part C: The change in potential energy is:
[
\Delta U = w L \sin(\theta)
]
These expressions show how the frictional force, the applied force, and the gravitational force contribute to the work and potential energy changes of the block as it moves up the incline.