37.8k views
2 votes
How much work did the movers do (horizontally) pushing a 41.0-kgkg crate 10.6 mm across a rough floor without acceleration, if the effective coefficient of friction was 0.60?

User Kangaswad
by
5.1k points

1 Answer

6 votes

Answer:

W=2.60 J

Step-by-step explanation:

Given that

m = 41 kg

d= 10.6 mm

Coefficient of friction = 0.6

Friction force ,Fr=μ m g

Fr= 0.6 x 41 x 10 N ( take g= 10 m/s²)

Fr= 246 N

We know that work done by a force given as

W = F . d

Now by putting the values

W= 246 x 10.6 x 10 ⁻³ J

W=2.60 J

Therefore the work done by movers in the horizontal direction will be 2.60 J.The mover does the work against the friction force.

User Wonderflame
by
4.7k points