A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J.What is the spring constant in this case?
A)10 000 N/m
B)5 000 N/m
C)1 200 N/m
D)No answer is correct.

Question

Quizplus · Accepted Answer

The potential energy stored in a spring is given by the equation:$$U = \frac{1}{2}kx^2$$where:* U is the potential energy stored in the spring (in joules)* k is the spring constant (in newtons per meter)* x is the displacement of the spring from equilibrium (in meters)In this case, we are given that the potential energy stored in the spring is 72.0 J and the displacement of the spring from equilibrium is 12.0 cm (or 0.12 m). Substituting these values into the equation above, we get:$$72.0 J = \frac{1}{2}k(0.12 m)^2$$Solving for k, we get:$$k = \frac{2(72.0 J)}{(0.12 m)^2} = 10,000 N/m$$Therefore, the spring constant in this case is 10,000 N/m.

A Hooke's Law Spring Is Compressed 12