Question

A block of mass 0.252 kg is placed on top of a light, vertical spring of force constant 4 875 N/m and pushed downward so that the spring is compressed by 0.109 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? (Round your answer to two decimal places.)

Answer 1

m = mass of the block = 0.252 kg

k = spring constant = 4875 N/m

d = spring compression = 0.109 m

g = acceleration by gravity = 9.81 m/s²

The elastic potential energy (PEe) in the compressed spring is:

PEe = k × d² / 2

PE = (4875 N/m) × (0.109 m)² / 2

PE = 28.959 J

When the block reaches its highest position, all that energy is converted to gravitational potential energy, so

PEe = m g h

or

h = PEe / (m g)

h = (28.959 J) / [(0.252 kg) × (9.81 m/s²)]

h = 11.7142 m

h = 11.71 m

Note that I used 9.81 m/s² for gravity.