a 2.2 kg ball is dropped from 3m above vertical spring whose spring constant is 450 n/m. a) how far will it compress the spring?

Question

b.) what velocity will the ball have when it leaves the spring again?

Answer 1

A) This is a conservation of energy problem, so E(i) = E(f). I'll pick our point of origin to be the spring's equilibrium height-- this makes the gravitational potential energy of the ball easy to calculate. We now identify what types of energy are present at the instant before the ball is dropped and at the instant where the spring is compressed and the ball's velocity is 0. Our conservation of energy equation becomes
mgh(i) = .5k(s-s(eq))^2 - mgs, where the second potential energy is negative because when the spring compresses, the ball is below our point of origin. s(eq) is the spring's equilibrium position, which is 0 because it is our origin. Move all terms to one side of the equation and quadratic formula.

B) Now pick your initial position to be the spring's maximum compression and final to be the instant the ball loses contact with the spring (which will be at the spring's equilibrium position). Our conservation of energy equation is:
.5k(s-s(eq)) + mgs = .5m[v(f)-v(i)]^2, where s is what we solved for in part A. The ball is initially at rest, so v(i) is 0. Solve for v(f).

Answer 2

Think you're reading too much into this Alex.
Simple mgh = 1/2kx^2, solve for x
b) 1/2kx^2 = 1/2mv^2