# Springs, elastic potential energy, kinetic energy

If a ball with some kinetic energy collides with a spring, the ball doesn't lose its kinetic energy in an instant, right? it loses kinetic energy as the spring gains potential elastic energy. Right?

So now I'm wondering. If I have a spring on the ground, and If I let the block falls from some meteres to collide with the spring on the ground. How can I calculate the maximum compression of the spring ?

I was thinking like this: The block has an Initial potential energy, that it's going to convert to kinetic energy, and at the time that the block collides with the spring the whole kinetic energy would transfer to the spring, so I could calculate all about the spring doing this:

say X is the initial potential energy of the BLOCK. Total Energy= Kinetic energy of the spring + Elastic Potential energy of the spring (=) X= Kinetic energy of the spring + Elastic Potential energy of the spring

But now I'm wondering if it wouldn't be more accurate to say that X= Kinetic energy of the spring + Elastic Potential energy of the spring - Kinetic energy of the ball !!

What do you think?

• We usually take the spring to be massless, so it has no kinetic energy. I think you need to work on clarifying what you are trying to say. – garyp May 26 '16 at 18:07
• I think you have to decide whether it's a ball or a block. If you drop a ball onto a stiff spring, the ball could bounce off without compressing the spring. (Technically that would be an elastic collision.) If you drop a block onto an easily compressed spring, then you're expecting an inelastic collision, i.e. the block stays in contact with the top of the spring. In that case, some of the energy will be converted to heat. So depending on your parameters, the heat loss may or may not be significant, and the kinetic energy of the spring may or may not be significant. – user3386109 May 26 '16 at 18:11
• @user3386109: that is not useful and quite confusing to the OP. Even a stiff spring will compress somewhat. That the cause of its elastic response. We're also assuming a Hookean spring: no heat is involved then. – Gert May 26 '16 at 19:15
• A massless, Hookean spring will convert the falling ball's kinetic energy to spring potential energy, until the ball momentarily stops. The restoring force of the spring will then start accelerating the ball upwards, until the spring's potential energy has been fully converted back to ball kinetic energy. You need to look up potential energy of a Hookean spring. – Gert May 26 '16 at 19:20
• @Gert You don't seem to understand the difference between an elastic response and an elastic/inelastic collision. It's clear from the question that the OP fully understands the theory behind a Hookean spring, and wants to go beyond that. – user3386109 May 26 '16 at 19:32

## 1 Answer

The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it and at some time the maximum compression is reached.

At maximum compression the block is momentarily stationary (between moving down and back up again) and the spring should be stationary as well (though I expect there would be some vibrations). My point is initially and at maximum compression there is no kinetic energy. Overall the initial gravitational energy is converted to elastic potential energy.

With a massive spring it becomes a bit more involved. The spring will compress to some extent under its on weight and as it is compressed its centre of mass will move down reducing its gravitational potential energy.

• ok, I get that. You understood the setup. I was trying to get an expression to describe the movement. So, from the moment that the block hits the spring, the exoression will be: TOTAL ENERGY = INITIAL GRAVTATIONAL POTENTIAL ENERGY= KINETIC ENERGY + ELASTIC POTENTIAL ENERGY. Considering the spring is massless. – Vitor Aguiar May 26 '16 at 21:26
• Don't forget the GRAVITATIONAL POTENTIAL ENERGY term in your second sum. What makes this problem a bit tricky is picking a reference level for zero gravitational energy. If you measure block's initial height from the top of the uncompressed spring then, as the spring is compressed, the block ends up with negative GRAVITATIONAL POTENTIAL ENERGY. – M. Enns May 26 '16 at 21:35
• oh, yes, that's what I was afraid of. So if the block is at some HEIGHT H above the spring , and we consider the top of the spring the reference level for zero gravitational energy, TOTAL ENERGY = GRAVTATIONAL POTENTIAL ENERGY+ KINETIC ENERGY + ELASTIC POTENTIAL ENERGY, where TOTAL ENERGY would be EQUAL TO THE INITIAL GRAVITATIONAL POTENTIAL ENERGY of the block ? – Vitor Aguiar May 26 '16 at 21:57
• Ya, that sounds right. Basically MgH = 1/2kx^2 + (-Mgx) where x is the amount of spring compression. – M. Enns May 27 '16 at 0:38