Question

A block of mass 0.244 kg is placed on top of a light, vertical spring of force constant 4 975 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise

Answer 1

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 (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