A 0.20-kg object is oscillating on a spring with a spring constant of k = 15 N/m.What is the potential energy of the system when the object displacement is 0.040 m,exactly half the maximum amplitude?
A)zero
B)0.006 0 J
C)0.012 J
D)2.5 J

Question

Quizplus · Accepted Answer

The potential energy stored in a spring is given by the formula:

U = (1/2) k x^2

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

When the object is at half of the maximum amplitude, its displacement is 0.040 m / 2 = 0.020 m. Therefore, the potential energy of the system is:

U = (1/2) k x^2 = (1/2) * 15 N/m * (0.020 m)^2 = 0.012 J

Therefore, the correct answer is C.

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