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3 votes
A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pushing the crate if the effective coefficient of friction was 0.50

User GoldenJoe
by
8.2k points

1 Answer

10 votes

Answer:

2,321.62Joules

Step-by-step explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone = 2,321.62Joules

Hence the amount of work done is 2,321.62Joules

User Jesse James
by
7.4k points
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