# Homework Help: Finding The Compression of a Spring

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1. Mar 21, 2015

### Astreiks

1. The problem statement, all variables and given/known data
"A coil spring has a force constant of k = 4.0 lb/in. When the spring's axis is inclined at an angle 30 degrees from the horizontal, a W = 2.0 oz ball is propelled to a height of 6.0 ft. By how much must the spring have been compressed initially? (1 lb = 16 oz)

2. Relevant equations
*PE = Potential Energy
*KE = Kinetic Energy

PEi = mgh
KEf = 0.5mv^2

F = ma
F = -kl

3. The attempt at a solution

So, I set PEi = KEf

mgh = 0.5mv^2
v^2 = mgh / 0.5m
v = SQRT(2gh)
v = SQRT(2 * 32.2 ft/s^2 * 6 ft)
v = 19.7 ft/s

Now, I'm just really confused. I don't know how to go farther than this. I'm figuring that Vf = 0 ft/s (at the topmost point) but I don't know much more than that. I'm assuming that acceleration = gravitational constant, but I'm not really sure where that gets me. In the end, finding the velocity seems kind of useless. Unless you can just set F = ma = -kx, but I'm not sure whether I can actually do that, seeing how the spring is at an angle...

Someone tried to explain this to me and told me that I'm calculating this all wrong because I'm not taking into account horizontal vs. vertical PE and KE, so I'm even more utterly lost. Any help would be appreciated.

2. Mar 21, 2015

### haruspex

Your calculation of v looks ok, but you need to be clear exactly what speed this represents.
As a projectile rises, what happens to its horizontal velocity? What can you say about vertical and horizontal velocities at the highest point?

Depending on how hard the question is intended to be, there is a subtlety to be considered. It doesn't say whether the height is measured from the compressed position or the relaxed spring position. To keep it simple, assume the first option.

3. Mar 21, 2015

### ME_student

Use the energy theorem equation.

T=KE
V=PE
U=WORK

T1+V1+U=T2+V2

4. Mar 21, 2015

### haruspex

That will certainly be needed, but Astreiks isn't quite ready to apply that yet.