# If you push something and it doesn't move does it have energy?

1. Aug 3, 2014

### needingtoknow

If you push something and it doesn't move does it have energy? So if I push something to apply a force, that requires energy right, but if the object like a building for example, doesn't move where does the energy go?

Another way I was looking at it was if a ball collides with another stationary ball and the collision is completely elastic, then the first moving ball will hit the stationary ball and stop and the target ball will move away with the same velocity as the first one. But let's take the same moving ball and let's say it collides with a stationary building, but instead it bounces back because it obviously cannot give up all its momentum and kinetic energy to the target (the building) like it did in the example with the stationary ball. So why exactly does the ball bounce back then?

Last edited: Aug 3, 2014
2. Aug 4, 2014

### haruspex

Muscles burn energy merely in tensing, so even though the object does not move, some work is done within the muscle. It turns to heat.
That's only true if the masses are the same. If the stationary ball was lighter then they will both continue forwards. If the stationary ball is the more massive then the first ball will bounce back somewhat.
This is the extreme case of the stationary ball being the more massive. Hardly any energy is transferred to it, nearly all going into the bounce back of the moving ball.

3. Aug 4, 2014

### rude man

Momentum and kinetic energy are still conserved, but you have to include the entire Earth in the picture.

Let m = mass of your ball with v1 and v1' its before-&-after velocity
M = mass of Earth with v2 and v2' its before-&-after velocity
Obviously, M >> m.

The expression
m v1 + M v2 = m v1' + M v2'

still holds. It's just that v1' = -v1 almost exactly (elastic collision assumed), while v2 = v2' = 0 almost exactly. I'm assuming perfectly rigid connection between the building and Earth. The center of mass of the ball and the Earth never moves, so as the ball moves to the right, say, the Earth must shift a tiny bit to the left to maintain the c.m. all the time. After the collision the opposite takes place, but still the c.m. does not move.

As haruspex points out, it's even more true with energy transfer. That's because kinetic energy is proportional to v2. So the slight change in v2 is even much tinier when squared, and it's even much more true that almost all kinetic energy stays with m.

4. Aug 4, 2014

### needingtoknow

Thank you both for your replies! Haruspex Im almost there in terms of understanding but one last question.

"This is the extreme case of the stationary ball being the more massive. Hardly any energy is transferred to it, nearly all going into the bounce back of the moving ball."

For this line, so the ball is carrying this kinetic energy. It hits the wall, at this point where is the energy because the ball is for this fraction of a second not moving because it is changing to the opposite direction. Where is the energy gone at the moment?

5. Aug 4, 2014

### Staff: Mentor

The ball has become slightly deformed, and this means that strain energy has been temporarily stored in the ball. This energy is released as the ball regains kinetic energy.

Chet

6. Aug 5, 2014

### needingtoknow

Thanks to everyone who posted! It makes sense now