Homework Help: Book dropped on spring conservation of energy

1. Mar 25, 2012

EEintraining

1. The problem statement, all variables and given/known data

A spring of negligible mass has force constant K = 1700 n/m

A book of mass 1.30 kg is dropped from a height of .7m above the top of the spring. Find the maximum distance the spring will be compressed.

2. Relevant equations

PEi + KEi = PEf + KEf

PEgrav = mgy
PEspring = 1/2Kx2
KEbook = 1/2Mv2

V=\sqrt [V2initial + 2 8 (9.8) *(.7)] = 3.70405 m/s approx
3. The attempt at a solution

I have tried this various ways without success. I think that I have the equation incorrect. I have:

PEi + KEi = PEf + KEf

PEgrav i + PEspring i + KEbook i = KEbook f + PEgrav f + PEspring f

mgy + 0 + 0 = 1/2Mv2 + 0 + 1/2Kx2
1.3 * 9.8 * .7 = 1/2 * 1.3 * 13.72 (which is v^2) + 0 + 1/2 * 1700 * x^2
8.918 = 8.918 + 850 x^2 which gives me 0

I know that I am missing something here

The first part of this question asks :
How far must the spring be compressed for an amount 3.50 J of potential energy to be stored in it?

Which I solved by using Uel = 1/2Kx2 but for some reason it is not working for the second part. The book is exerting a force of mgy 1.3*9.8*.7 on the spring and I should be able to set that equal to 1/2Kx2 but it was not correct.

Can someone tell me what I am missing here? Sorry this is long I tried to be detailed with what I have so far.

2. Mar 25, 2012

Staff: Mentor

Don't forget that the spring compresses, which involves a lowering of the mass and a further change in gravitational PE. Hint: Measure the gravitational PE from the lowest point.

3. Mar 25, 2012

EEintraining

Do you mean on the second side of my equation? I choose zero to be where the book impacts the spring. So the spring compresses at that point would I have gravitational PE that is negative due to my coordinate system? Also, for the kinetic energy, do I need to change this somehow to account for the fact that it is still moving as it hits the spring?

4. Mar 25, 2012

Staff: Mentor

Nothing wrong with that.
Exactly!
Why bother to worry about this intermediate point? (Unless they ask about it, of course.) Instead, just compare the initial point (the moment the book is released) and the final point (when the spring is compressed to the maximum).

5. Mar 25, 2012

EEintraining

So I would then need to change the right side of my equation as well?

mgy would equal 1.3*9.8* (.7+x)

6. Mar 25, 2012

EEintraining

crap.... what i did was also incorrect

if i taking the point where the book is dropped to the point where the spring is fully compressed i took:
x is distance spring is compressed

1.3*9.8(.7+x) = 8.918 (kinetic) + 1.3*9.8(-x)+1/2 Kx^2
12.74(.7+x) = 8.918 + -12.74x + 850x^2
8.918 + 12.74x = 8.918 + -12.74x + 850x^2
25.48x = 850x^2 divide by 850x
x = .03 which is wrong as well

7. Mar 25, 2012

Staff: Mentor

(1) If you are measuring gravitational PE from the point where the book meets the spring, what would be the gravitational PE at the initial point where the book is released?
(2) What's the KE when the spring is maximally compressed?

8. Mar 25, 2012

EEintraining

OK I think I get it... i was probably over thinking it. So as soon as it hits the spring it it no longer has gravitational potential because it is on the spring and converting it into potential energy for the spring?

Thanks for your help!

9. Mar 25, 2012

Staff: Mentor

If you measure the gravitational PE from the initial position of the spring (before being compressed), then the mass will have zero gravitational PE when it just hits the spring.

But who cares? Why are you worrying about that intermediate point? Just compare the total energy at these two points:
(1) The starting point, just as the book is released.
(2) When the spring is fully compressed.

10. Mar 25, 2012

EEintraining

Thanks for clearing that up for me!