How Does Potential Energy Change During a Free Fall Without Air Resistance?

[DrummingAtom]

Homework Statement

A mass, m, is released from height, h, off a cliff. No air resistance.

(a) How much Work is done on the mass by the gravitational force during the fall?

(b) What is the change in PE of the mass-Earth system during the fall?

(c) If that PE is taken to be zero at the ground, what is its value when the mass is released?

(d) If, instead, the PE is taken to be zero at the release point, then what is its value when the mass reaches the ground?

Homework Equations

W= F*d
PE = m*g*y

The Attempt at a Solution

(a) W= 9.8*m*h, understand that one.

(b) Change in PE is PE_final-PE_initial, which is 0 - 9.8*m*h.

(c) This is where I get a bit lost.. I would think its the same answer as (b) because I was already assuming that the PE_final is going to be zero when on the ground anyway. I peeked at the answers and this one = 9.8*m*h Positive...

(d) Same problem here.. I'm thinking that PE_final-PE_initial would be 9.8*m*h - 0. But again it's reversed than how I'm getting it. This answer is -9.8*m*h

Any help is appreciated, thanks.

 

[Tusike]

Potential energy shows how much work is done by the force when the object is taken from point A to the zero-point, O.
I think you have b) correct.
As for c), yes it is the same as b). In b) you had PE(final) - PE(initial) which is 0 - mgh, from which PE(initial) = mgh.
As for d), apply the definition: how much work is done by gravity when an object is taken from the ground to the release-point? Answer: -mgh, so that's the potential energy on the ground if you put zero at the release point. So, -mgh - 0 = -mgh.

When you have signs mixed up, I advise that you take a look at the definitions, it seems they weren't really clear.