Question

When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.78 cm.

(a) What is the force constant of the spring?
N/m

(b) If the 2.00-kg object is removed, how far will the spring stretch if a 1.00-kg block is hung on it?
cm

(c) How much work must an external agent do to stretch the same spring 9.00 cm from its unstretched position?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Force constant of the spring = k

Mass of the first object = m₁ = 2 kg

Extension of the spring due to the first object = X₁ = 2.78 cm = 0.0278 m

m₁g = kX₁

(2)(9.81) = k(0.0278)

k = 705.755 N/m

Mass of the second object = m₂ = 1 kg

Extension of the spring due to the second object = X₂

m₂g = kX₂

(1)(9.81) = (705.755)X₂

X₂ = 0.0139 m

X₂ = 1.39 cm

Work done to stretch the spring by 9 cm = W

X = 9 cm = 0.09 m

W = kX²/2

W = (705.755)(0.09)²/2

W = 2.858 J

a) Force constant of the spring = 705.755 N/m

b) Amount the spring will stretch due to the 1 kg block = 1.39 cm

c) Work done by an external agent to stretch the spring by 9 cm = 2.858 J