# Find the maximum distance the spring is compressed.

1. Homework Statement
A block of mass 2.0kg is dropped from height h=70cm onto a spring of spring constant k=1960 N/m. Find the maximum distance the spring is compressed.

2. Homework Equations
KE initial + PE initial = KE final + PE final
KE = 1/2(mass)(velocity)^2
PE = mgh or, when dealing with a spring, PE = 1/2 (k)(compression distance)^2

3. The Attempt at a Solution

The initial velocity is zero, so the initial KE is zero
The final height is zero so the final PE is zero

2kg(9.8)(.7m) = (1/2)(1960)x^2

where x is the distance the spring is compressed
solve for x.... x = .118m
but it is not the right answer....

Any help is appreciated,
Thanks,
A

Related Introductory Physics Homework Help News on Phys.org
Hint: before the block falls on the spring, the gravitational potential energy is not mgh since the spring compresses a bit.

but I thought if I wanted to calculate the total amount the spring compressed, the amount of energy at the very end of the fall (the instant before hits the spring-before it compresses) should equal the amount of energy the block has when it stops on the spring.... I am very confused!
Should KE at the end of the fall = (1/2)(1960)x^2 ?

That would be true if the spring somehow compressed without the top end of the spring actually being compressed downward, which is impossible.

I'm sorry~ I'm still not quite sure I understand:
so as the block hits the spring:
1/2m(vinitial)^2 + 1/2k(xinital)^2 + mgh = 1/2k(xfinal)^2
??
if x final = 0, then the whole thing = 0... which doesn't work out very well (square root of a negative number...)

Set your 0 point so that at the point where the spring is compressed at its lowest point, that is your h = 0. So then how far above the lowest point is the block before you let it go?

x + .7 meters?

Now you can move on.

Thank you!!
(they should make a monument to super physics people like you)
:P