Book dropped on spring conservation of energy

[EEintraining]

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 = 1/2Kx²
KE_book = 1/2Mv²

V=\sqrt [V²_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 = 1/2Mv² + 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 U_el = 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.

 

[Doc Al]

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.

 

[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?

 

[Doc Al]

I choose zero to be where the book impacts the spring.

Nothing wrong with that.

So the spring compresses at that point would I have gravitational PE that is negative due to my coordinate system?

Exactly!

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?

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).

 

[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)

 

[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

 

[Doc Al]

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

(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?

 

[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!

 

[Doc Al]

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?

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.

 

[EEintraining]

Thanks for clearing that up for me!