Question

A relaxed spring of length 0.19 m stands vertically on the floor; its stiffness is 1270 N/m. You release a block of mass 0.5 kg from rest, with the bottom of the block 0.7 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? m

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the block = m = 0.5 kg

Force constant of the spring = k = 1270 N/m

Length of the spring = L₀ = 0.19 m

Height of the block above the floor = H₀ = 0.7 m

Initial height of the block from the top of the spring = H

H = H₀ - L₀

H = 0.7 - 0.19

H = 0.51 m

Compression of the spring when the block comes momentarily to rest = X

The initial potential energy of the block is converted into the potential energy of the spring as the block comes to rest on the spring.

mg(H + X) = kX²/2

(0.5)(9.81)(0.51 + X) = (1270)X²/2

2.502 + 4.905X = 635X²

635X² - 4.905X - 2.502 = 0

X = 0.0667 m or -0.059 m

Distance cannot be negative.

X = 0.0667 m

Length of the spring when the block comes momentarily to rest = L

L = L₀ - X

L = 0.19 - 0.0667

L = 0.1233 m

Length of the spring when the block comes momentarily to rest on the compressed spring = 0.1233 m