- #1
Dark Visitor
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Ernie is holding a 2 kg ball at rest at the opening (top) of a 10 m deep well. He releases the ball so that it falls into the well. Assume the bottom of the well represents zero potential energy position. Immediately before the ball hits the bottom, the sum of it's kinetic energy and potential energy is:
* 0
* 196 J
* -196 J
* 392 J
* -392 J
The only 2 equations I see usable for this are:
Umg = mgy
K = .5mv2
What I was thinking of doing was getting the kinetic energy, but I realized there is no velocity given, so I am not sure if I should find that, or if I don't need to bother with it. We know the initial velocity is zero.
I also thought that using the Potential energy equation, we would get:
Umg = (2 kg)(9.8 m/s2)(10 m) = 196 J
but I am unsure because I don't know if the 10 m is negative because the well is under ground, or if it stays positive.
I could use some help starting this problem. I need it by tomorrow night, but I would like to get it done a.s.a.p. Thanks for any help.
* 0
* 196 J
* -196 J
* 392 J
* -392 J
The only 2 equations I see usable for this are:
Umg = mgy
K = .5mv2
What I was thinking of doing was getting the kinetic energy, but I realized there is no velocity given, so I am not sure if I should find that, or if I don't need to bother with it. We know the initial velocity is zero.
I also thought that using the Potential energy equation, we would get:
Umg = (2 kg)(9.8 m/s2)(10 m) = 196 J
but I am unsure because I don't know if the 10 m is negative because the well is under ground, or if it stays positive.
I could use some help starting this problem. I need it by tomorrow night, but I would like to get it done a.s.a.p. Thanks for any help.