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Quick gravitational potential energy question...?
For a ball dropped on earth, the initial gravitational potential energy is 50.0 J
how would you find initial gravitational PE for the ball on the moon where g = 1.67 m/s^2
I know that PEgrav = m*g*y
So I know that PEgrav on Earth = 50.0 J
and I know g ... but aren't I missing both m and y?
So how could I continue?
Once I know this I should be okay.. thanks!
2 Answers
- DavidLv 71 decade agoFavorite Answer
If I were reading this problem I would have to assume that "y" is the same for both situations (you're right, you couldn't solve it any other way!)
So we have two equations:
50 = m(9.8)y
U = m(1.67)y
We can solve this by making a ratio:
U/50 = m(1.67)y / m(9.8)y. You can see that m and y will cancel here.
U/50 = 1.67/9.8
Multiplying by 50 on both sides yields a moon potential energy of 8.52 joules.
Hope this helped!
- RickBLv 71 decade ago
I think they're assuming that, in the moon case, you are using the SAME ball at the SAME height as the earth case.
So in other words, the "m" and the "y" are the same in both equations.
PE_earth = m(g_earth)y = 50J
PE_moon = m(g_moon)y
Divide 2nd equation by 1st:
PE_moon/PE_earth = g_moon/g_earth
(notice how "m" and "y" cancelled out.)