Question

A spring stretches out by 8cm when a 0.2 kg weight is placed on it. How much does the spring stretch by when a 0.5 kg weight is placed on it?

Answer 1

m₁ = mass hanged initially = 0.2 kg

x₁ = initial stretch in the spring = 8 cm = 0.08 m

k = spring constant of the spring

the weight of the mass hanged is balanced by the spring force by the spring hence

Spring force = weight of mass hanged

k x₁ = m₁ g

k (0.08) = (0.2) (9.8)

k = 24.5 N/m

m₂ = mass hanged later = 0.5 kg

x₂ = final stretch in the spring = ?

k = spring constant of the spring = 24.5 N/m

the weight of the mass hanged is balanced by the spring force by the spring hence

Spring force = weight of mass hanged

k x₂ = m₂ g

(24.5) x₂ = (0.5) (9.8)

x₂ = 0.2 m = 20 cm