- #1

EEintraining

- 31

- 0

## Homework Statement

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.

## Homework Equations

PE

_{i}+ KE

_{i}= PE

_{f}+ KE

_{f}

PE

_{grav}= mgy

PE

_{spring}= 1/2Kx

^{2}

KE

_{book}= 1/2Mv

^{2}

V=\sqrt [V

^{2}

_{initial}+ 2 8 (9.8) *(.7)] = 3.70405 m/s approx

## The Attempt at a Solution

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

PE

_{i}+ KE

_{i}= PE

_{f}+ KE

_{f}

PE

_{grav i}+ PE

_{spring i}+ KE

_{book i}= KE

_{book f}+ PE

_{grav f}+ PE

_{spring f}

mgy + 0 + 0 = 1/2Mv

^{2}+ 0 + 1/2Kx

^{2}

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 U

_{el}= 1/2Kx

^{2}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 1/2Kx

^{2}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.