Kinetic and Potential Energy of Falling Objects

[submar1ney]

Hi. If i drop a stone from a height, i understand its Gravitational Potential Energy will decrease and its kinetic enrgy increases.

Am i correct in saying that when the stone hits the ground and comes to rest, this kinetic energy is now 'stored' potential energy?

Regards

Ian

 

[berkeman]

Hi. If i drop a stone from a height, i understand its Gravitational Potential Energy will decrease and its kinetic enrgy increases.

Am i correct in saying that when the stone hits the ground and comes to rest, this kinetic energy is now 'stored' potential energy?

Regards

Ian

No, when it hits the ground, the kinetic energy is converted to heat and sound energy, and goes into deforming the ground (which may be a form of stored energy, depending on the nature of the ground).

 

[submar1ney]

No, when it hits the ground, the kinetic energy is converted to heat and sound energy, and goes into deforming the ground (which may be a form of stored energy, depending on the nature of the ground).

So if it was solid ground, with no deformation, it would all be converted to sound and heat?
So, is gravitational potential energy only 'stored' when the stone is raised again?

Cheers

Ian

 

[PatPwnt]

Yes, when you raise something from the ground, you are giving it potential energy.

 

[berkeman]

So if it was solid ground, with no deformation, it would all be converted to sound and heat?
So, is gravitational potential energy only 'stored' when the stone is raised again?

Cheers

Ian

If it was like a steel ball hitting a steel floor, the ball would likely bounce. With little energy lost at the bounce except to sound, the ball would rise back up almost to its starting height, before heading back down for another bounce.

Yes, gravitational PE is increased when you do the work to lift something against the gravitational force.

 

[submar1ney]

Thanks guys, that's cleared things up for me. I understand now.

Regards

Ian

 

[submar1ney]

Just one more quickie guys please.

I'm calculating the 'Loss of GPE' of this falling stone. I see 'loss of GPE' the same as 'change in GPE', therefore have used the formula: 'delta'GPE= m g 'delta' h where the 'delta' h is the difference in height at 2 different points in time. Is this correct?

Cheers

Ian

 

[berkeman]

Just one more quickie guys please.

I'm calculating the 'Loss of GPE' of this falling stone. I see 'loss of GPE' the same as 'change in GPE', therefore have used the formula: 'delta'GPE= m g 'delta' h where the 'delta' h is the difference in height at 2 different points in time. Is this correct?

Cheers

Ian

Oops, then this would be homework/coursework. I've moved the thread to Homework Help. Please take care to post homework/coursework questions in the Homework Help forums. Even if it is for self-study, it still should be placed in HH.

You are correct in your use of the formula \Delta {PE} = mg \Delta {h}

Can you tell us how this relates to the work done? What is the definition of work done in moving an object?

 

[submar1ney]

OOOps, sorry about that.

Well, when work is done on an object, any energy transferred is stored and is calculated by mg \Delta {h}

Work done on an object by, in this case, gravitational force will cause kinetic energy to increase.

Cheers

Ian

 

[berkeman]

OOOps, sorry about that.

Well, when work is done on an object, any energy transferred is stored and is calculated by mg \Delta {h}

Work done on an object by, in this case, gravitational force will cause kinetic energy to increase.

Cheers

Ian

True, but what I was looking for was that work is force through a distance. W=Fd. In your current gravitation-based situation, the work is mg \Delta {h} because the gravitational force is constant pointed down, and you are applying a constant force to raise the object (assuming that you are not accelerating it as well -- that would take extra force above mg). Makes sense?

 

[submar1ney]

Ah yes, much clearer now. Many thanks

Regards

Ian