Question

How much potential energy is stored in a spring with a spring constant of 27 N/m if it is stretched by 16 cm

Answer 1

Answer:

PE = 0.35 J

Explanation:

Elastic potential energy in a spring is defined by PE = (1/2)kx^2, where k is the spring constant and x is the distance compressed/stretched. Converting 16 cm to 0.16 m, we get

PE = (1/2) * 27 * (0.16^2)

PE = 0.35 J