Question

Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process? (The asnwer is 0 J, please explain!)

Answer 1

W = F*dx*cos(a)

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)

=> W = -F*x

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

Also, total work is the sum of all 45 passes by the sandpaper.

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)*(2.6N)*(0.92)*(0.12)
W = -12.9168 Joules