A 0.250 kg block on a vertical spring with a

Question

spring constant of 4.09 × 10
3
N/m is pushed
downward, compressing the spring 0.0900 m.
When released, the block leaves the spring
and travels upward vertically.
The acceleration of gravity is 9.81 m/s
2
.
How high does it rise above the point of
release?
Answer in units of m

Answer 1

Call height at spring unextended = 0
then height at bottom = -.09
height at top = h
(we want h +.09)
velocity is 0 at bottom and again at top so we are talking all potential energy here

At bottom total Pe = (1/2)k (.09)^2 - m g(.09)

At top total Pe = (1/2) k h^2+ m g h

those are the same so
(1/2)k h^2 + m g h=(1/2)k(.09)^2-.09 m g

solve for h, add .09