Contradicting conclusions on falling bodies.

  • Thread starter rishch
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  • #1
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I was recently thinking about potential energy and I came to two conclusions which contradict each other-

1)When a body is falling from a height, the force of gravity is acting on it in the downwards direction. There is displacement in the downwards direction. So as the force and the displacement are in the same direction, some amount of positive work is done on it.

2)When the same body is falling, the potential energy of the body is converted into kinetic energy and the total energy of the ball remains constant. So that means that no work is being done on it.

So in one case there is positive work while in the other no work. How? What's wrong in my reasoning.Please keep the answers as simple as possible because I'm still in secondary school, and won't be able to understand a highly complicated answer. So please keep it as simple as possible.
 

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  • #2
CAF123
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By the work-energy theorem, [tex] \frac {1}{2} mv_f^2 - \frac {1}{2} mv_i^2 = W, [/tex]if kinetic energy has increased, then positive work must have been done.
 
  • #3
Doc Al
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1)When a body is falling from a height, the force of gravity is acting on it in the downwards direction. There is displacement in the downwards direction. So as the force and the displacement are in the same direction, some amount of positive work is done on it.
Sure. So the kinetic energy increases. (Note that we do not use the concept of gravitational PE here.)

2)When the same body is falling, the potential energy of the body is converted into kinetic energy and the total energy of the ball remains constant. So that means that no work is being done on it.
Here we account for the effect of gravity and the work it does by using the concept of gravitational PE.

Using the concept of gravitational PE already includes the effect of gravity, so by counting both the work done by gravity and gravitational PE you end up counting it twice.

Work done by all forces (including gravity) = ΔKE

Work done by all forces (except gravity) = ΔKE + ΔPE
 
  • #4
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So you're saying that we cannot include the work done by the force of gravity because this work is already included in the potential energy? And for the last sentence where you say "except gravity" that means you are using the concept of PE here?
 
  • #5
Doc Al
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So you're saying that we cannot include the work done by the force of gravity because this work is already included in the potential energy? And for the last sentence where you say "except gravity" that means you are using the concept of PE here?
Right.

Another way to think of it is this:

Work done by all forces (except gravity) + Work done by gravity = ΔKE
Work done by all forces (except gravity) - mgΔh = ΔKE
Work done by all forces (except gravity) = ΔKE + mgΔh
 
  • #6
AlephZero
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So you're saying that we cannot include the work done by the force of gravity because this work is already included in the potential energy?
That's basically right, but I would put it a different way: "The force of gravity" and "gravitational potential energy" are really two different descriptions of the same thing. So when you work out the energy change, you shouldn't include the same effect twice.

If a body moves along any path you choose between two fixed points A and B, the amount of work done by the force of gravity is always the same. That means it is possible to define "gravitational potential energy" which depends only on the position of the body in space, and the difference GPE(A) - GPE(B) measures the amount of work done moving between the two points.

If you are still in secondary school you might have only come across a simpler formula that the difference in GPE = mgh, where h is the vertical distance between the two points. That is only approximately true close to the surface of the earth, but the general idea of GPE apples in any situation.
 
  • #7
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Thanks, I think understand now. And yes, mgh is the only formula I've learnt.
 

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