Maximum Mass for Spring-Induced Jump

[tmiddlet]

This is not a homework/coursework problem, but I came across it and wanted to check my answer.

Homework Statement

As in the diagram, there are two blocks of mass M and m. The mass m is suspended above M by a spring of spring constant k.

Initially, the block m is pushed down with a force 3mg until the system reaches equilibrium, then the force is released.

What is the maximum value for M for which the bottom block will jump off the ground?

Homework Equations

F = kx
F = mg
U = \frac{1}{2}kx^2

The Attempt at a Solution

I think this is the right way to do it, but I'm not sure.

To find the equilibrium offset:
3mg = kx
x = \frac{3mg}{k}

Finding the potential energy due to the spring:
\frac{1}{2}kx^2 = \frac{9m^2g^2}{2k}

To find the maximum upward offset (d) of the spring after release:
\int_{-3mg/k}^{d} (-kx-mg)dx = -\frac{9m^2g^2}{2k}

Evaluating this returns : (If I did it right)
d = \frac{mg(\sqrt{7}-1)}{k}

If kx > Mg, then the block will lift off, so I end up with:
M \lt m(\sqrt{7}-1)


I'm just curious if I got this right. If anybody is willing to try the problem or check my logic, that would be great. Thank you!

 

[ehild]

Your logic works quite well, but:

To find the equilibrium offset:
3mg = kx
x = \frac{3mg}{k}

The block is pushed with a force of 3mg, but there is gravity, too, so the net force is 4mg.

To find the maximum upward offset (d) of the spring after release:
\int_{-3mg/k}^{d} (-kx-mg)dx = -\frac{9m^2g^2}{2k}

The integral is the work done by all forces between its initial and final position which is equal to the change of KE according to the work-energy theorem. d is the minimum upward displacement above the height of the relaxed spring where the spring force is Mg/k. As that, the velocity of the upper block is zero there. So the integral has to be zero.

ehild

 

[tmiddlet]

Ok, I see.

Easy mistake:
x = \frac{4mg}{k}

It seems everything else I did was unnecessary, I don't need to deal with the potential energy.

So evaluating the integral:
\int_{\frac{-4mg}{k}}^{d} (-kx - mg) dx = 0

Yields:
d =\frac{2mg}{k}
M \lt 2m


This makes sense, Thanks!