Question

How much work did the movers do (horizontally) pushing a 130-kgkg crate 10.5 mm across a rough floor without acceleration, if the effective coefficient of friction was 0.50?

Answer 1

6.6885Joules

Step-by-step explanation:

Wordone is expressed as the product of the moving force acting on the body and the distance.

Workdone = Force * distance

According to Newton second law;

Fm - Ff = ma

Since acceleration is zero;

Fm - Ff = 0

Fm = Ff

Fm is the moving force

Ff is the frictional force

Since Ff = ηR

Fm = Ff = ηR

η is the coefficient of friction

R is the reaction

Get the moving force:

Fm = ηR

Fm = ηmg

Fm = 0.50(130)(9.8)

Fm = 637N

Recall that distance = 10.5mm = 0.0105m

Get the work done;

Work done = 637 * 0.0105

Work done = 6.6885Joules

Hence the movers did 6.6885Joules of work.