A 0.20-kg object is oscillating on a spring with a spring constant of k = 30 N/m.What is the potential energy of the system when the object's displacement is 0.040 m,exactly half the maximum amplitude?
A) 0.96 J
B) 0.024 J
C) 0.012 J
D) zero

Question

Quizplus · Accepted Answer

The potential energy of the spring-mass system can be calculated using the formula:

U = 1/2 kx^2

where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

When the object's displacement is exactly half the maximum amplitude (which is 0.080 m), x = 0.040 m.

Plugging in the given values:

U = 1/2 (30 N/m) (0.040 m)^2
U = 0.024 J

Therefore, the potential energy of the system is 0.024 J, which corresponds to choice B.

A 020-Kg Object Is Oscillating on a Spring with a Spring