The Student Room Group

Work Energy and power - Help(again)

can some one explan the process behind conversion of energy so for example
a ball falls 3.2 meters to the ground. using energy consevation only what is its final speed as it hits the surface.

we have been told to do it like this but i dont understand it
Loss of GPE = Gain in KE
mg^h = 1/2m(v*v)

KEY ^=Delta
(v*v) = squared
I can't really see what you're stuck on tbh...

are you happy that the ball has velocity at the instant it first begins to touch the floor and hasn't started to compress yet?
Reply 2
Original post by connerdom
can some one explan the process behind conversion of energy so for example
a ball falls 3.2 meters to the ground. using energy consevation only what is its final speed as it hits the surface.

we have been told to do it like this but i dont understand it
Loss of GPE = Gain in KE
mg^h = 1/2m(v*v)

KEY ^=Delta
(v*v) = squared


Hi there. The gravitational potential energy will be converted into kinetic energy as it falls so you can make use of mgh = 1/2mv^2. Remember energy is never lost, its just transferred therefore the energy the ball has due to its position in the earth's gravitional field (height) will be used up as the ball falls. In this case the M's cancel out leaving you with gh = 1/2v^2. You can rearrange this equation to get v and subsititue in the correct values. I hope that helps.
(edited 9 years ago)
Reply 3
I've had a quick read of this and I highly suggest that you do some reading on this topic as most of what you've stated here is inaccurate! Unfortunately I think this will only confuse the OP more so :frown:
Reply 4
Your conclusion that the change in GPE of the object is equal to the change in KE is correct, however the rest of it was quite incorrect, I could see the intention but it confused even me and I'm doing a physics degree!

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