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## Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a total dista

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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

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Physics
3 years
2021-08-20T07:30:49+00:00
2021-08-20T07:30:49+00:00 1 Answers
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## Answers ( )

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