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?

Question

asked Jun 2, 2020 109k views

1 Answer

Kit Barnes · Answer 1 · 2020-06-06T11:19:44+0000

Answer:

Work done, W = 10195.92 Joules

Step-by-step 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 :

$W=\mu mg* d$

W=0.6* 170* 9.8 * 10.2

W = 10195.92 Joules

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