Question

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.

Answer 1

Part A:
`\( W_f = - \mu w L \cos(\theta) \)`

Part B:
`\( W = F L \cos(\theta) - \mu w L \sin(\theta) \)`

Part C:
`\( \Delta U = w L \sin(\theta) \)`

Step-by-step explanation:

In Part A, the work done by the force of friction
`(\( W_f \))` can be expressed as
`\( W_f = -f d \cos(\theta) \)`, where f is the force of friction. Since
`\( f = \mu w \)` and the displacement d is L , the expression becomes
`\( W_f = - \mu w L \cos(\theta) \)`, where the negative sign indicates the opposing direction of the force of friction.

For Part B, the work done by the applied force W is the sum of the work done by the applied force and the work done against friction. The applied force contributes
`\( F L \cos(\theta) \)` to the work, and the force of friction contributes
`\( -\mu w L \sin(\theta) \)`, resulting in the expression
`\( W = F L \cos(\theta) - \mu w L \sin(\theta) \)`.

In Part C, the change in potential energy
`(\( \Delta U \))` is related to the change in height of the block. As the block is pushed up the incline, the change in height is
`\( L \sin(\theta) \)`, and since potential energy is given by U = mgh , where m is mass and g is the acceleration due to gravity,
`\( \Delta U = w L \sin(\theta) \)`.

Therefore, the final expressions for the three parts are
`\( W_f = - \mu w L \cos(\theta) \)`,
`\( W = F L \cos(\theta) - \mu w L \sin(\theta) \)`, and
`\( \Delta U = w L \sin(\theta) \)`.

Answer 2

The work done by friction is -μwLsinθ, the work done by the applied force is FLcosθ, and the change in potential energy is wLsinθ.

Step-by-step explanation:

Part A: The work done on the block by the force of friction as it moves a distance L up the incline can be calculated using the equation Wf = -μwLsinθ. The negative sign indicates that the frictional force is acting opposite to the motion of the block. Therefore, the work done by friction will be negative.

Part B: The work done by the applied force of magnitude F can be calculated using the equation W = FLcosθ. The cosine of the angle θ is used because the applied force is acting parallel to the incline.

Part C: The change in potential energy of the block after it has been pushed a distance L up the incline can be calculated using the equation ΔU = wLsinθ. The change in potential energy is equal to the work done against gravity.