Spring energy

Tags:
1. May 23, 2015

transmini

So this sounds homework question but I promise its not. At least its not mine. I saw it on a website because someone made it funny because of what they answered (drew an elephant if you know the one I'm talking about)

Anyway, so the problem was a block of mass m = 5 kg falls starting at rest at h = 5 m down a curved frictionless ramp into a spring with spring constant k = 100 N/m. Find the distance it compresses x and find the height the block comes back up to.

I understand using gravitational potential energy is equal to spring/elastic potential energy at the max compression to find x, but what I don't quite get is finding the max height the block comes back up to. I'm finding online that its the same as the starting height because all the potential energy in the spring is pushed back in to the block and since the ramp is frictionless. But wouldn't the spring still oscillate hinting that it has energy? If it has energy, wouldn't this energy had to have been taken from the kinetic energy of the block? Otherwise how does the spring get the energy to oscillate and still push the block back up to initial height?

2. May 23, 2015

Kyouran

The spring's oscillatory motion means that it would have kinetic energy too, but this then implies that it should have mass. The problem is if you consider it to be a massless spring, the differential equation describing this oscillatory motion would become kx = 0, meaning that it doesn't move at all. In a real world scenario, the spring would have mass, and the mass in the spring is what keeps its oscillation going, described by the solution to mx"+kx = 0 when there is no friction, and then the block would indeed not reach its initial height again because energy is transferred to the spring.

3. May 23, 2015

transmini

That makes sense that its being assumed as massless and that it wouldn't ACTUALLY behave like that. So would the spring be leeching its energy from the kinetic energy of the block then? And while the answer makes sense, is there a way to show without differential equations? I'm just starting my degree and don't do anything with those until this fall.

4. May 23, 2015

Kyouran

If the spring had a mass, yes it would be leeching energy from the block, although the correct physical way to say this is to say that the block does work on the spring, reducing its own energy and increasing the energy of the spring. Conservation of energy would still hold if you consider the system being the block + the spring then, as the total energy of the block and spring together would still equal the potential energy of the block at the start. The problem however is that the law of conservation of energy will only allow you to solve for one single unknown, and you cannot thus find the solution using just energy conservation in that case. The energy stored in the spring would be given by E:

$$m_b g (h_{initial} - h_{final} ) = E$$

which is therefore a second unknown in that equation. What you can do is use conservation of linear momentum along with conservation of energy, which should allow you to solve for the remaining unknowns.

As for the oscillatory motion of the spring...if you know that the spring has eventually stored an amount of energy equal to E (after the block has separated from the spring), conservation of energy on the spring alone will give you another differential equation:

$$\frac{kx^2}{2}+\frac{m_s v^2}{2} = E$$

where v = dx/dt. I can tell you that the solution to this is $$x(t) = \sqrt{\frac{2E}{k}} \sin(\sqrt{\frac{k}{m_s}} t + \phi)$$ which you can check by doing the differentiation and substitution in the equation above.

EDIT: Also, I'm not sure whether you can show that the spring oscillates without going into differential equations.

Last edited: May 23, 2015
5. May 23, 2015

sophiecentaur

It's even harder to consider what would happen in subsequent descents of the mass. It could strike the spring in any phase of its self oscillation and either increase or decrease the spring's oscillatory energy. The problem is in the realm of Coupled Oscillators but the analysis could be a bit hard because of the impulsive situation when the mass hits the spring each time. Including the losses would make things even worse.