1. Oct 24, 2005

### NutriGrainKiller

My physics class is intro more or less (it's calc based; 111 NOT 101). I'm a bit confused when it comes to springs..specifically, finding the maximum compression a spring will have.

Here are the given variables in a problem in my textbook:

*spring has negligible mass
*force constant(k) = 1600n/m
*spring is placed virtically with one end on the floor
*1.2kg book is dropped from a height of .80m above the top of the spring

maximum distance the spring will be compressed?

what confuses me specifically are the signs (+/-) and the distances involved. Here is what I am thinking:

I need to use K1 + U1 = K2 + U2 (k1 = zero)

now when it comes to potential energy, i don't know whether to use MGY or 1/2kx^2..can someone point me in the right direction? thanks!

2. Oct 24, 2005

### Staff: Mentor

Good. (What will K2 be when the the spring is maximally compressed?)
Since the height of the book changes, you need gravitational PE; since the spring is compressed, you need spring PE. Use both!

3. Oct 24, 2005

### NutriGrainKiller

So are you saying that K2=zero as well? i guess that would make sense

So..I need U1 (spring PE) + U2 (grav PE) = U1 (spring PE) + U2 (grav PE)

-or-

1/2MV1^2 + mgy1 = 1/2MV2^2 + mgy2 ?

EDIT: Another question; I'm having trouble differentiating between the Y's..where should I solve for Y and where should I plug in .8?

Last edited: Oct 24, 2005
4. Oct 24, 2005

### Staff: Mentor

If K2 wasn't zero, the spring could not be fully compressed--the book would still be moving.
That will work.
This won't work--you left out spring PE.
Call the amount that the spring compresses "x". Then pick a reference point for gravitational PE (may as well call the unstretched position of the spring to be y = 0) and set up the equation. Give it a shot.

5. Oct 24, 2005

### NutriGrainKiller

sorry - should have 1/2KX1^2 + MGY1 = 1/2KX2^2 + MGY2

x is how much the spring is compressed. 0 is where the spring is at rest (so -x would be the compression)

1/2(1600)x^2 + (1.2)(9.8)(.8) = 1/2(1600)x^2 + (1.2)(9.8)(.8)

....it's the same on both sides would cancel out..

6. Oct 24, 2005

### Staff: Mentor

Initially the spring is uncompressed, so x1 = 0. Initially the book is at y1 = 0.8; but where is it when the spring is compressed by amount "x"?

7. Oct 24, 2005

### NutriGrainKiller

...-x?

so here's what i have now:

mgy1 = 1/2kx2^2 + mg(-x)

or

9.408 = 800x2^2 - 11.76x <--looks like quadratic

8. Oct 24, 2005

### Staff: Mentor

Good! Now you're cooking. Solve that equation.

9. Oct 24, 2005

### NutriGrainKiller

somehow i usually tend to screw the quadratic equation up, but here is what i got:

(11.76 +/- sqrt(30243.9))/1600

which ends up being: .116m or 11.6cm

the answer in the book was 12cm, i don't think the sig figs required that type of rounding

10. Oct 24, 2005

### Staff: Mentor

One of those solutions should be negative. That one can be discarded as physically irrelevant.

(Your initial data was only good to two sig figs; so your answer should be rounded off accordingly.)

11. Oct 24, 2005

### NutriGrainKiller

so how do i get the answer if this equation is discarded?

12. Oct 24, 2005