Question

an ideal spring has a spring constant of 15 N/m. The spring is stretched 6 cm. What is the potential energy of the spring?

Answer 1

Answer:

P.E = 0.027 Joules.

Explanation:

Given the following data;

Spring constant, k = 15N/m

Extension, e = 6cm to meters = 6/100 = 0.06m

To find the potential energy;

P.E = ½ke²

Substituting into the equation, we have;

P.E = ½*15*0.06²

P.E = 7.5 * 0.0036

P.E = 0.027 Joules