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?

W=2.60 J

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.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .