Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a total distance of 0.15 m. Kevin pushes the sandpaper against the surface with a normal force of 1.8 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the kinetic frictional force during the sanding process

Respuesta :

W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules

The expression for the work allows to find the result for the work done by the friction force is:

         [tex]W_{total}[/tex] = - 11,178 J

Work is defined by the scalar product of force and displacement.

          W = F. d

Where bold indicates vectors, W is work, F is force, and d is displacement.

We can develop the scalar product, remaining

           W = F d cos θ

where θ is the angle between the force and the displacement.

The friction force is the macroscopic sum of the interaction between the molecules of the two surfaces and always opposes the movement, it can be represented for solid bodies by the equation.

           fr = μ N

.

Where fr is the friction force,  μ is the friction coefficient and N is the normal.

Indicate that the normal force is 1.8 N, the displacement is 0.15 m and the coefficient of friction is 0.92.

As the friction opposes the movement, it has an angle of 180º with respect to the displacement, see attached.

       

We substitute.

          W =  μ N d cos 180

          W = - 0.92 1.8 0.15

          W = -0.2484 J

Each time the sandpaper passes, it performs this work, at 45 times the total work is

         [tex]W_{total}[/tex] = n W

         [tex]W_{total}[/tex] = 45 (-0.2484)

         [tex]W_{total}[/tex] = -11,178 J

In conclusion using the expression for work we can find the work of the friction force is:

         [tex]W_{total}[/tex] = - 11,178 J

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