A Hooke's law spring is compressed 12.0 cm from equilibrium,and the potential energy stored is 36.0 J.What compression (as measured from equilibrium)would result in 100 J being stored in this case?
A) 20.0 cm
B) 14.1 cm
C) 13.6 cm
D) The correct answer is not given.

Question

Quizplus · Accepted Answer

The potential energy stored in a spring is given by:$$U = \frac{1}{2}kx^2$$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 its equilibrium position (in meters)In this case, we are given that the potential energy stored in the spring is 36.0 J when the spring is compressed 12.0 cm from equilibrium. We can use this information to find the spring constant:$$36.0 J = \frac{1}{2}k(0.12 m)^2$$$$k = \frac{2(36.0 J)}{(0.12 m)^2} = 2000 N/m$$Now, we can use the spring constant to find the compression that would result in 100 J being stored in the spring:$$100 J = \frac{1}{2}(2000 N/m)x^2$$$$x = \sqrt{\frac{2(100 J)}{2000 N/m}} = 0.2 m = 20.0 cm$$Therefore, the correct answer is A) 20.0 cm.

A Hooke's Law Spring Is Compressed 12