Question

when an object is moved from rest at point a to rest at point b in a gravitational field, the net work done by the field depends on the mass of the object and

Answer 1

Step-by-step explanation:

Remember that work is a transition function - not a state function. You only do work over a distance, not at a point. So when using W=FΔx

W=∫rfriFdx

Let's choose the positive axis outwards from Earth (force is negative −Fgrf<ri

W=∫rfri(−Fg)dx=[−(−GMx)]rfri=[GMx]rfri=GMrf−GMri

rf<riGMrf>GMriW

Or we can try with another choice of axis, e.g. from the starting point of the falling object and inwards towards Earth (force is positive +Fgrf>ri

W=∫rfriFgdx[−GMx]rfri=(−GMrf)−(−GMri)=GMri−GMrf

In this case ri<rfGMri>GMrf

This can also be looked at from pure energy.

Gravitational potential energy is:

U=∫Fdr=−GMr

Work done by Earth on something falling is:

W=ΔK=Ui−Uf=−GMri−(−GMrf)=GMrf−GMri

Again work W

A sign rule-of-thumb is always the formula: W=Fr

Gravity therefor always does positive work on a falling object. But was rising like a weatherballoon, the work done by gravity would be negative. Negative work just means that your efforts don't really work; the object still moves in another direction than in which you are pushing/pulling.