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.

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 - =

       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

The normal reaction prevents the block from sinking into the inclined plane

Work done by friction on the block is  μ·L·cos(θ)(F₁·sin(θ) - m·g)

The reason above expression for work done is correct is presented as follows:

The given parameters are;

The magnitude of the force acting on the block = F₁

Coefficient of kinetic friction between the plane and the block = μ

Work done by the force of friction =

The angle of inclination of the plane = θ

Distance up the incline the block moves = L

Required:

The work done on the block by the force of friction,

Solution:

Normal reaction, N = (m·g - F₁·sin(θ))·cos(θ)

Fictional force, = -μ·(m·g - F₁·sin(θ))·cos(θ) (opposite in direction to the applied force)

Work done = Force × Direction of the force

Work done on the block by friction, as the block moves up the inclined plane a distance, L is = × L

∴ =  -μ·(m·g - F₁·sin(θ))·cos(θ) × L

Work done on the block by friction, as the block moves up the inclined plane a distance, L, =  -μ·(m·g - F₁·sin(θ))·cos(θ) × L= μ·L·F₁·cos(θ)·sin(θ) - μ·m·g·L·cos(θ) = μ·L·cos(θ)(F₁·sin(θ) - m·g)

Work done on the block by friction = μ·L·cos(θ)(F₁·sin(θ) - m·g)

Learn more about motion on an inclined plane here:

link

6 cos 34, 6 sin 34 trig

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