A stationary mass m = 1.3 kg is hanging from a spring of spring constant k = 1200 N/m.You raise the mass a distance of 10 cm above its equilibrium position.How much has the potential energy of the mass-spring system changed?
A)1.3 J
B)6.0 J
C)7.3 J
D)12 J
E)13 J

Question

Quizplus · Accepted Answer

The change in potential energy is given by the formula ΔPE = 0.5 * k * x^2, where x is the displacement from equilibrium. Here, x = 0.1 m, so ΔPE = 0.5 * 1200 * (0.1)^2 = 6 J.

A Stationary Mass M = 1