A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 52.0 J.What is the spring constant in this case?
A)7 220 N/m
B)3 610 N/m
C)72.2 N/m
D)0.722 N/m
E)No answer is correct.

Question

Quizplus · Accepted Answer

The potential energy stored in a spring is given by the formula PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium. Rearranging for k gives k = 2 * PE / x^2. Substituting PE = 52.0 J and x = 0.12 m, we get k = 2 * 52.0 / (0.12)^2 = 7,222.22 N/m, which rounds to 7,220 N/m.

A Hooke's Law Spring Is Compressed 12