# Spring Hooke's law

1. Jan 31, 2008

### zvee_y

[SOLVED] Spring Hooke's law

1. The problem statement, all variables and given/known data
The innerspring mattress is held up by 23 vertical springs, each having a spring constant of 5000N/m. A 44 kg person jumps from a 1.94m platform onto the innersprings. Assume the springs were initially unstretched and that they stretch equally. Determine the Stretch of each of the springs.

3. The attempt at a solution
I use the equation mgh=(44)(9.8)(1.94) to find the potential energy. To find the stretch displacement, is it correct if I use Us=(1/2)kx^2 and substitute Us=mgh, k=5000 as given, and find x is the stretch of each springs?

2. Jan 31, 2008

### Dick

It is if you multiply Us=(1/2)kx^2 by the number of springs and then equate to mgh.

3. Feb 1, 2008

### zvee_y

Ok. So i did mgh=44*9.8*1.94=836.528J
Us=(1/2)*5000*x^2*23=836.528J. Find x from there to get stretch displacement. But the answer wasn't right. Did I do anything wrong in here?

4. Feb 1, 2008

### Sourabh N

As springs stretch(or compress), the person's potential energy also changes by mgx, where x is the stretch in the spring.

5. Feb 1, 2008

### zvee_y

but they give h=1.94m. how is it a stretch?

6. Feb 1, 2008

### Sourabh N

The height is from spring when it is unstretched. When the person jumps on spring, the spring compresses and reduces in length, so the person's effective change in potential energy is mgh +mgx. (Hope i'm clear now)

7. Feb 1, 2008

### zvee_y

oh ok i understand what you are explaining now. So to find x, we set it mgh+mgx=(1/2)kx^2*23 ?

8. Feb 1, 2008

### zvee_y

can someone please help me with this questionn?? Im confused. :(

9. Feb 2, 2008

### Sourabh N

Didn't you get the correct answer using the formula in #7 ?

10. Feb 2, 2008

### zvee_y

Well i got it right. But i just don't understand how is it mgh+mgx ??

11. Feb 2, 2008

### Sourabh N

Look, when the person is about to touch the bed (i.e springs), the change in PE is mgh; now as the springs compress, the person (along with spring) goes down (try to visualize the situation), and its PE further decreases.

Some other PF guy may give you a better explanation if my explanation is not clear enough.

12. Feb 2, 2008

### Dick

There is nothing wrong with Sourabh N's explanation. When the springs are at their maximum compression, there is no kinetic energy. Only potential. So the change in the gravitational potential (over a distance of h+x) and potential in the springs (over a distance x) must be equal.

13. Feb 2, 2008

### zvee_y

hehe Thank you. I think i got it. Just never think about that equation you know. Thanks a lots. Yes now i even understand more through DIck's explanation. mg(h+x)=(1/2)kx^2*23 hhehe...i totally got it now.

Last edited: Feb 2, 2008