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Gravitational Potential Energy and Work Energy Theorem |
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| Apr18-12, 06:34 AM | #1 |
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Gravitational Potential Energy and Work Energy Theorem
Just a little confused, when work is done on an object it's energy increases right? (i.e. Work done on object = Change P.E + Change K.E + Work done by object)
So how come when gravitational force does work on an object(i.e. it falls) the potential energy decreases instead. So its like Work done by gravitational force = Decrease in P.E + Increase in K.E = 0? Edit : Okay, thinking about it probably because of action-reaction pair with the Earth? So work done on body = Decrease in P.E + Increase in K.E + work done by body? |
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| Apr18-12, 08:18 AM | #2 |
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When you use that particular form of the work-energy theorem--which includes a gravitational PE term--the work done means the work done by forces other than gravity. The effect of gravity is already included in the PE term. So, using that form of the theorem, you'd never include work done by gravity as part of the "work done" term. 0 = Change PE + Change KE As the object falls the PE decreases while the KE increases. |
| Apr18-12, 09:12 AM | #3 |
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So in case of mechanics, net work done(external energy transfer)(if I include gravitational force) never changes potential energy because work causes change in velocity not position (Am I right to say this)? Sorry if it seems illogical I'm trying to connect the dots. |
| Apr18-12, 11:59 AM | #4 |
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Gravitational Potential Energy and Work Energy TheoremWork done by all forces = ΔKE If gravity is the only force acting, then: Work done by gravity = ΔKE -mgΔh = ΔKE Rearranging: 0 = mgΔh + ΔKE = ΔPE + ΔKE If you include the work done by gravity and the ΔPE term in your statement of the Work-Energy theorem, you'll get the wrong answer because you'll be counting gravity twice. |
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