## Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a to

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

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3 years 2021-08-20T07:23:05+00:00 1 Answers 5 views 0

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

W = 12.96 J