Here is an illustration to ease understanding:
Angle inclined plane: θ = 20.0°
spring of force constant k = 525 N/m
block: m = 2.71 kg
distance of block from spring: d = 0.330 m
speed of block: v = 0.750 m/s
By what distance is the spring compressed when the block momentarily comes to rest? (I'm going to call it 'x' in this problem.)
The answer should be x = 0.143 m
Ok, I'm not quite sure how many of these are relevant or if I'm missing some, but I'll just give some that I know.
PEs = (1/2)kx2 (x = distance spring is compressed)
PEg = mgh
KE = (1/2)mv2
F = ma
U + KE = 0 (U = any potential energy? All potential energies? I'm not quite positive what is going on here.)
The Attempt at a Solution
Okay. I'm not quite sure how to go about this, so I'll just start doing what I seem to know.
Let's start with force. It doesn't seem inherently relevant to the problem, but why not?
Okay, so force in the x direction seems to be: mgsinθ = ma. So we get a = gsinθ = (9.8)(sin 20°) = 3.35198 m/s2.
Force in the y direction seems to be: N - mgsinθ = 0. So N = mgsinθ = (2.71)(9.8)(sin 20°) = 9.08337 N.
Ok. That was probably not relevant. Let's do work now.
Using the numbers above, I get KE = (1/2)(2.71)(.750)2 = 0.76219 J.
PEg = (2.71)(9.8)h. Okay, I'm not quite sure if h should just be d or if I have to factor in the slope. I'm thinking that slope probably does have to be included, so h is 0.330(sin 20°), so PEg is 2.99751 J.
So it seems like PEg + PEs + KE = 0. Then it must be that (1/2)kx2 + mgh + (1/2)mv2 = 0... However, solving for x gives me the square root of a negative number, I must be interpreting U + KE = 0 wrong. Where do I go from here? Thanks, I know this is long.