Gravitational Potential Energy and Work Energy Theorem

[hqjb]

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?

 

[Doc Al]

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]

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.

 

[Doc Al]

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.

 

[asteriatic]

I can help clarify the concept of gravitational potential energy and the work-energy theorem for you. First, let's define these terms:

- Gravitational potential energy is the energy that an object possesses due to its position in a gravitational field. This energy is related to the object's mass, height, and the strength of the gravitational field.

- The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (energy of motion) plus the change in its potential energy.

So, when work is done on an object, its energy does indeed increase. This is because the work done on the object is converted into kinetic energy and potential energy. However, when an object falls due to the force of gravity, its potential energy decreases. This may seem counterintuitive, but it is because the work done by the gravitational force is being converted into kinetic energy, not potential energy.

To understand this, let's look at the example of a ball being dropped from a height. As the ball falls, the force of gravity is doing work on it, converting its potential energy into kinetic energy. This is why the ball's potential energy decreases. However, the ball's kinetic energy increases, and according to the work-energy theorem, the work done by the gravitational force is equal to the change in kinetic energy plus the change in potential energy. In this case, the change in potential energy is negative (decreasing), but the change in kinetic energy is positive (increasing), so the total work done by the gravitational force is still positive.

You are correct in thinking that this is due to the action-reaction pair between the object and the Earth. The Earth's gravitational force is acting on the object, but the object is also exerting an equal and opposite force on the Earth. The work done by the object on the Earth is equal and opposite to the work done by the Earth on the object, so they cancel each other out.

In summary, when an object falls due to the force of gravity, its potential energy decreases, but its kinetic energy increases. This is because the work done by the gravitational force is being converted into kinetic energy, according to the work-energy theorem. I hope this helps clarify the concept for you.