Question

A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is μ.What is the total work WfricWfricW_fric done on the block by the force of friction as the block moves a distance LLL up the incline? Express the work done by friction in terms of any or all of the variables μμmu, mmm, ggg, θθtheta, LLL, and F1F1F_1.

Answer 1

Answer:

W = (F1 - mg sin θ) L,   W = -μ  mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis  

       N - $W_{y}$ =

       N = W_{y}

X axis

       F1 - fr - Wₓ = 0

       fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     cos θ = W_{y} / W

      Wₓ = W sin θ

      W_{y} = W cos θ

We substitute

      fr = F1 - W sin θ

Work is defined by

        W = F .dx

        W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

         

        W = -fr x

        W = (F1 - mg sin θ) L

Another way to calculate is

         fr = μ N

         fr = μ W cos θ

the work is

         W = -μ  mg cos θ L