Question

How much work did the movers do (horizontally) pushing a 43.0-kg crate 10.4 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60

Answer 1

The crate is in equilibrium. Newton's second law gives

∑ F (vertical) = n - mg = 0

∑ F (horizontal) = p - f = 0

where

• n = magnitude of the normal force

• mg = weight of the crate

• p = mag. of push exerted by movers

• f = mag. of kinetic friciton, with f = 0.60n

It follows that

p = f = 0.60mg = 0.60 (43.0 kg) g = 252.84 N

so that the movers perform

W = p (10.4 m) ≈ 2600 J

of work on the crate. (The total work done on the crate, on the other hand, is zero because the net force on the crate is zero.)