A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 36.0 J.What is the spring constant in this case?
A) 10,000 N/m
B) 5000 N/m
C) 1200 N/m
D) The correct answer is not given.

Question

Quizplus · Accepted Answer

The potential energy stored in a spring is given by $ PE = \frac{1}{2} k x^2 $. Solving for $ k $, we have $ k = \frac{2 \times PE}{x^2} $. Substituting $ PE = 36.0 \, \text{J} $ and $ x = 0.12 \, \text{m} $, we get $ k = \frac{2 \times 36.0}{0.12^2} = 5000 \, \text{N/m} $.

A Hooke's Law Spring Is Compressed 12