Question

A relaxed spring of length 0.15 m stands vertically on the floor; its stiffness is 1070 N/m. You release a block of mass 0.5 kg from rest, with the bottom of the block 0.6 m above the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring?

Answer 1

Answer:

x' = 0.085 m = 8.5 cm

Explanation:

The law of conservation of energy says that:

Potential Energy Stored in Spring = Loss in Gravitational Potential Energy of Block

(1/2)kΔx² = mgh

where,

k = stiffness of spring = 1070 N/m

Δx = change in length of spring = ?

m = mass of block = 0.5 kg

g = 9.8 m/s²

h = height of block above spring = 0.6 m - 0.15 m = 0.45 m

Therefore,

(1/2)(1070 N/m)Δx² = (0.5 kg)(9.8 m/s²)(0.45 m)

Δx = √[2(2.205 Nm)/(1070 N/m)]

Δx = 0.064 m

but,

Δx = x - x' = 0.15 m - x' = 0.064 m

x' = 0.15 m - 0.064 m

x' = 0.085 m = 8.5 cm