Is gravity a from of energy?

[brian334]

Gravity makes things move, does it have energy?

[Klaynos]

Gravity is one of the fundemental forces. Like all forces it has a potential energy associated with it:

 

[math]E=\frac {-GMm}{r}[/math]

[brian334]

If a 65 lb weight fell 10 feet, what would the impact speed be? And how much force/energy would it have?

[Klaynos]

This looks a lot like homework...

 

Well I don't work in imperial... So no numbers can I provide and even if I did, I wouldn't the point of homework is for you to work thew it yourself... but I shall help

 

You can use a simplified version of the equation I used above to work out the change in potential energy:

 

[math]\Delta E = mg \Delta h[/math]

 

Where m is the mass, g is the acceleration due to gravity, [math]\Delta E[/math] is the change in energy, and [math]\Delta h[math] is the change in height.

 

This energy has to be transferred into some other energy what could this be?

[brian334]

This is not homework.

[ydoaPs]

This looks a lot like homework...

 

Well I don't work in imperial... So no numbers can I provide and even if I did, I wouldn't the point of homework is for you to work thew it yourself... but I shall help

 

You can use a simplified version of the equation I used above to work out the change in potential energy:

 

[math]\Delta E = mg \Delta h[/math]

 

Where m is the mass, g is the acceleration due to gravity, [math]\Delta E[/math] is the change in energy, and [math]\Delta h[math] is the change in height.

 

This energy has to be transferred into some other energy what could this be?

I thought [math]\Delta E[/math] in your equation is the change in gravitational potential energy, not the change in total mechanical energy.

 

IIRC, in situations such as falling objects, KE₁+PE₁=KE₂+PE₂. If you choose your reference frame right, you can make some terms cancel. I'll leave that up to you.

[Klaynos]

I was using E for gravitational potential energy. Not best practise I'll take that but still perfectly valid. I call it that in the explanation before the equation... It will of course be exactly equal to the change in kinetic energy as shown from your equation of total energy.