Physics 1 Question?

Question

Hey! Please any help on this problem! I am really stuck on this one. Any help would be great.

A 0.250-kg block is placed on a light vertical spring (k = 5.00 x 10^3 N/m) and pushed downwards, compresing the spring 0.100 m.

After the block is released, it leaves the spring and continues to travel upwards. What height above the point of release will the block reach if air resistance is negligible?

Answer 1

This is an energy question. The total energy at the bottom when the spring is compressed equals total energy at top. All energy at bottom is elastic energy, all energy at top is gravitational energy. (luckily no kinetic energy at either point here because the block isn't moving.)
PEgrav = mgh where m =mass in kg
g = 9.8m/s2 and h is height above release.
PEelast = 1/2 k (delta x)squared
where delta x is the compression distance = .1m
and k is the spring constant given.
set these two equations equal and you have all variables except h.
as usual it is hard to show with this editor, but your text should have both equations.

Answer 2

I could be wrong and assume the wrong things, but couldn't you simply conserve the energy in the mass-spring system? Meaning: the energy stored in the spring at the depressed state would have to equal the difference in potential energy (due to height difference) at the final state?

So the energy stored in the spring is 0.5*k*dx^2 = 0.5*5000*0.1^2=25. That has to equal to the potential energy gained when the block reaches the new height with respect to the current height. So 25=m*g*dh=0.25*9.8*dh, giving dh=10.2m. That's 10.2m from the depressed height. When referenced to the height of the uncompressed spring, the block should reach 10.1m above that if the block is released.

That may be too simplistic though. I don't know if your instructor wants you to solve the equations of motion starting from forces. It's kind of a mess. I seriously doubt a physics 1 class would want you to solve the equations of motion of something like this.