Gravitational Potential Energy and Work Energy Theorem


by hqjb
Tags: energy, gravitational, potential, theorem, work
hqjb
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#1
Apr18-12, 06:34 AM
P: 39
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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Doc Al
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Apr18-12, 08:18 AM
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Quote Quote by hqjb View Post
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)
I think you mean: Work done on object = Change PE + Change KE

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.
So how come when gravitational force does work on an object(i.e. it falls) the potential energy decreases instead.
When an object falls, the only force acting is gravity. And since that's already included in the PE term, you'd set "work done on object (by forces other than gravity)" = 0. So:
0 = Change PE + Change KE

As the object falls the PE decreases while the KE increases.
hqjb
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Apr18-12, 09:12 AM
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Quote Quote by Doc Al View Post
I think you mean: Work done on object = Change PE + Change KE

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.

When an object falls, the only force acting is gravity. And since that's already included in the PE term, you'd set "work done on object (by forces other than gravity)" = 0. So:
0 = Change PE + Change KE

As the object falls the PE decreases while the KE increases.
I kinda get it so during free fall, the Change in P.E. is negative = Work done by gravitational force on body = Gain in K.E.? If i bring it over it becomes 0.

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.

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Apr18-12, 11:59 AM
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Gravitational Potential Energy and Work Energy Theorem


Quote Quote by hqjb View Post
I kinda get it so during free fall, the Change in P.E. is negative = Work done by gravitational force on body = Gain in K.E.? If i bring it over it becomes 0.

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.
Let me try again. The most basic form of the Work-Energy theorem is this:
Work 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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