As the spring returns to it's equilibrium position, it performs
1/2 (4975 N/m) (0.097 m)² ≈ 23 J
while the gravitational force (opposing the block's upward motion) performs
-(0.244 kg) g<em> </em>(0.097 m) ≈ -2.3 J
of work on the block. By the work energy theorem, the total work done on the block is equal to the change in its kinetic energy:
23 J - 2.3 J = 1/2 (0.244 kg) v² - 0
where v is the speed of the block at the moment it returns to the equilibrium position. Solve for v :
v² = (23 J - 2.3 J) / (1/2 (0.244 kg))
v = √((23 J - 2.3 J) / (1/2 (0.244 kg)))
v ≈ 44 m/s
After leaving the spring, block is in free fall, and at its maximum height h it has zero vertical velocity.
0² - (44 m/s)² = 2 (-g) h
Solve for h :
h = (44 m/s)² / (2g)
h ≈ 2.3 m
