Question

An elastic spring stores 45 J of potential energy when it is stretched by 2 cm. What is the spring constant?

Answer 1

PE = 45 J

x = 2 cm = 0.02 m

U= 1/2 k ( x)^2

Where:

U= energy stored

x = stretch

k = spring constant

Isolate k:

k = 2U/ x^2

k = ( 2 * 45 J) / (0.02)^2

k= 225,000 N/m