# Gravitational potential energy-help!

1. Mar 21, 2007

### a seeker

hi, I came across a question in my physics textbook that I didn't get, so I was hoping that you guys could help me out a bit.

Here's the question. It's about gravitational potential energy, by the way.

** On your desk you have N identical coins, each with a mass m. You stack the coins into a vertical pile to height y.
a) Approximately how much work, in terms of m, g, and y, must you do on the last coin to raise it from the desk to the top of the pile?
b) Approximately how much gravitational potential energy, in terms of m, g, N, and y, is stored in the entire pile?

of course, we have the equation E_g=mg delta y.

I was thinking that for a), delta y would be something like y/N(N-1), y/N being the height of one of the coins. It's because you have to consider that the height of the coins should not include the one that you are raising from the desk. Is it right? and as for b), I have no clue. Can it be zero, or no?

2. Mar 21, 2007

### robb_

Part a asks "approximately," you just need to consider how much work it takes to lift a mass up to the top of the stack.
Part b, each mass requires different amount of work since the final height changes, add the individual energies up.

3. Mar 21, 2007

### a seeker

well.. I am supposed to derive an equation for the situations using the variables given (like m, g, y, and N).

"Part a asks "approximately," you just need to consider how much work it takes to lift a mass up to the top of the stack."
-robb, i don't really understand what you are getting at. the work it takes to lift a mass up is indeed what I am asking, but how can you make an equation or a statement out of the variables?

and as for part b, why is there "a final height change"? can you be more specific? do we have to consider E_g of each coin, and not that of the whole pile (of coins)?

Last edited: Mar 21, 2007
4. Mar 21, 2007

### robb_

Well, you can get very technical for part a if you like and find the height of each coin since the n coins forms a stack of height y. Then the center of mass for the last coin is a distance y minus half the height of one coin. I do not think that is needed here though. You simply have a mass m that sits a height y above the zero of potential Gravitational PE of the top coin can be written in terms of these given quantities.
The second part is more interesting and involves a sum or integral, since the PE of each coin varies.

5. Mar 21, 2007

### a seeker

so you think gravitatial potential energy is simply E_g=mg delta y?

that's what i thought at first, but shouldn't we exclude the height of the coin that we are lifting up and make delta y= y/N(N-1)?
it's what I said at the beginning. *y/N=the height of one of the coins.

6. Mar 21, 2007

### robb_

I understand your concerns.
As long as the thickness of one coin is much smaller than the total height we can ignore the fact that the top coin, when placed, has its center of gravity at a point below the position y above the table, as that position is to the top of the coin.
But in case your curious, I still do not think that you have the change in height of the last coin correctly.
Just as a check: your delta y for the last coin is much smaller than the value y itself. That doesn't seem correct, right?, it should be slightly less than the value y.

* again the question says approximately*