...a 53.71 kg object falls off the top

Question

A 53.71 kg object falls off the top of 77.61 m height. How much work is required to lift the object back to its original position, assuming the lifting is down with a constant velocity.

Answer 1

I will assume instead that the object is lifted back UP at constant velocity.

The work required (without friction) is the change in gravitational potential energy, ΔPE = mgh = (53.71 kg) (9.8 m/s²) (77.61 m) = 4.1 x 10⁴ J. The sig figs are limited by the value of g; if you use g = 9.81 m/s², then the work can be written as 4.09 x 10⁴ J.