A rope is attached to a 50.0-kg crate to pull it up a frictionless incline at constant speed to a height of 3-meters. Note that the force of gravity has two components (parallel and perpendicular component); the parallel component balances the applied force and the perpendicular component balances the normal force.
Calculate the amount of work done upon the crate.
Answer: Wext = 1470 J
Start with TMEi + Wext = TMEf
KEi + PEi + Wext = KEf + PEf
KEi + 0 J + Wext = KEf + (50 kg) * (9.8 m/s/s) * (3 m)
(KEi = KEf since speed is constant. Thus, both KE terms can be eliminated from the equation.)
Wext = (50 kg) * (9.8 m/s/s) * (3 m) = 1470 J
But why? Isn't that the crate is moving with constant speed, which means net force is zero. So why does the total work done not equal zero?
thank you so much in advance!!!