# Homework Help: Need some help Springs

1. Jan 15, 2007

### godindisguise

A wingless fly of mass m sits on top of a platform of mass M supported by a spring of constant k. At time t the fly decides to jump upwards and by doing so, reaches a maximum altitude of 2d above the equilibrium point of the unloaded spring (i.e. the spring without the platform and fly). Assuming that the spring has no mass, find what is the maximum depth reached by the platform.

2. Relevant equations
F=-kx

3. The attempt at a solution
I have no idea how the fly jumping up a certain height has anything to do with the depth at which the spring originally was...the most simple answer I could come up with:

F = -kx
mg+MG = -kx
x = -g(m+M)/k

Thanks

2. Jan 16, 2007

### AlephZero

That looks correct for the initial position of the spring. The question is about what happens AFTER the fly jumps. What happens to the platform, after the fly jumps off it?

3. Jan 16, 2007

### godindisguise

Thanks..but REALLY need some more help...

4. Jan 17, 2007

### AlephZero

Imagine somebody made a video of this and played it backwards.

You see the fly falling from a given height onto the platform mounted on a spring. (Imagine its a big fly and a small platform). Does that remind you of any type of problem you have seen before? What equations would you use to solve it?

5. Jan 18, 2007

### skiboka33

I am a little confused by the question. Like the original post, I can't see how the fly's height would mean anything. If anything, wouldn't this just cause the spring to release slightly?

6. Jan 18, 2007

### tim_lou

after the fly jumps, the fly gains momentum. by the conservation of momentum, what is the momentum of the platform immediately after the fly jumps? (hint: consider the energy of the fly)

the following equation may be helpful:
$$KE=\frac{p^2}{2m}$$

from there, how can you find the maximum extension of the spring?

Last edited: Jan 18, 2007