Question

A 1 kg mass is attached to a vertical spring, stretching it by 10 cm from its equilibrium position. With the spring secured in the same stretched position, the 1 kg mass is removed and replaced with a 0.50 kg mass. If the spring is then released, how high will the 0.50 kg mass rise above the point where it was released?

Answer 1

Case 1 :

M = mass attached to spring = 1 kg

x = stretch of the spring = 10 cm = 0.10 m

k = spring constant of the spring = ?

Using equilibrium of force in vertical direction

k x = Mg

k (0.10) = 1 (9.8)

k = 98 N/m

case 2 :

m = new mass attached = 0.50 kg

h = height gained above the point of release

Using conservation of energy

Gravitational potential energy gained by new mass = Spring potential energy

mgh = (0.5) k x²

(0.50) (9.8) h = (0.5) (98) (0.10)²

h = 0.1 m