Question

How much work did the movers do (horizontally) pushing a 170-kg crate 10.2 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60?

Answer 1

Answer:

Work done, W = 10195.92 Joules

Explanation:

Given that,

Mass of the crate, m = 170 kg

Distance, d = 10.2 m

The coefficient of friction, $\mu=0.6$

Let W is the work done by the mover. It is given by in terms of coefficient of friction as :

$W=\mu mg\times d$

$W=0.6\times 170\times 9.8 \times 10.2$

W = 10195.92 Joules

So, the work done by the mover is 10195.92 Joules. Hence, this is the required solution.