# Mass dropped on a spring

1. Dec 8, 2005

### Leomar

Hi, hope you can help me with finding out what im doing wrong on this problem.
It goes as follows:
A massless, vertical spring of force konstant k is attached at the bottom to a platform of mass mp, and at the top to a massless cup.
The platform rests on a scale. A ball of mass mb is dropped into the cup fro a negligible height. What is the reading on the scale a) when the ball momentarily comes to a rest with the spring compressed.
And b) same as a), but with the ball dropped from a height h.
For problem a), I set that the ball would have potentional energy=0 when at rest, and thus that the work done by the spring must equal mbgx.
1/2kx^2=mbgx
=>
x=2mbg/k
Using Hookes law, the force exerted by the spring on the ball is 2mbg.
So the spring exerts the same force on the platform, and the scale should show F=mpg+2mbg
Which according to the solutionsguide is correct. (Though I might have made a mistake here anyway)
For problem b), I tried using the same method, setting that
1/2kx^2=mbg(x´+h) (Where x=2mbg/k, from problem a))
=>
x=(2mbg(2mbg/k+h)/k)^(1/2)
rearranging
x=2mbg/k(1+hk/2mbg)^(1/2)
Which gives the solution
F=mpg+2mbg(1+hk/2mbg)^(1/2)
This is where the solutionsguide is disagreeing with me.
A hint or two would be very much appreciated.

2. Dec 8, 2005

### Fermat

Thanks for posting your work, leomar. It makes it easier to give an answer.

x is different from x'

x is the compression when the ball is just placed in the cup.

When the ball is then dropped from a height h above the cup before striking it, then it has additional energy which compresss the spring further and so x' > x.

3. Dec 9, 2005

### Leomar

I realise that the the x in a) and the x in b) are different values, but I did think that the different amount of work needed to be done by the spring would be equal to mgh. Where h is the extra height.
I might have been confusing by using x in equation a), then in be refering to it, but with a different name, sorry about that.

But it really boils down to, am I doing anything wrong in assuming that the work done by the spring in situation b), 1/2kx^2, equals to the work done in situation a), mbgx' (Note that this x' is the lenght the spring was compressed in a), not how much it will be compresses in b)) plus the extra energy gained from the heigh h, which would equal mbgh.
1/2kx^2=mbg(x´+h)
(Where x=2mbg/k, from problem a))

Hope that clears it up.

4. Dec 9, 2005

### Fermat

The work done is different in either case.
The WD in b) is greater than the WD in a) because the spring is comprressed more in case b).
For a) WD = (1/2)kx² = mgx
For b) WD = (1/2)kx'² = mg(x' + h)
To find the new force on the scale, find F = mpg + kx'
You seem to have been be using x for part a),
In my explanation above, I have used x as the compression for part a) and x' as the compression for part b).

5. Dec 9, 2005

### Leomar

Ah, now I see, thank you very much for you help.
Should be able to figure it out now

Edit: Yep, that did it. Again, thanks.

Last edited: Dec 9, 2005