A 0250 kg block is placed on a light vertical spring k 540

A 0.250 kg block is placed on a light vertical spring (k = 5.40 10³ N/m) and pushed downward, compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upward. What height above the point of release will the block reach if air resistance is negligible?
1 m

Solution

M=0.300kg

K=5.40*10^3=5400N/m

X=0.100m

H=?

From the conservation of energy ,Ki=Kf=0

Ug=0 at the time of release

(Ug+Us)i=(Ug+Us)f

0+Usi=Ugf+0

1/2kx^2=mgh

H=kx^2/2mg

=5400*0.100^2/2*0.300*9.8

H=9.18 m--answer