A 6.95 kg bowling ball is dropped 2.20 m onto a large spring. If the bowling ball compresses the spring by 0.55 m, what must the spring constant of the spring be?

Answer 1

total drop = 2.2+ .55 = 2.75 m
drop in gravity Potential energy =
= 6.95 * 9.81 * 2.75
= increase in spring Pe
= (1/2)k x^2 = .5 *k * .55^2
so

k = (6.95*9.81*2.75)/(.5*.55*.55)