Question

A stretch spring has an elastic potential energy of 35 J when it is stretched 0.54m. What is the spring constant of the spring?

Answer 1

Answer:

240.1 N/m

Explanation:

Applying the formula for the potential energy stored in a stretched spring,

E = 1/2ke².................... Equation 1

Where E = potential energy, k = spring constant, e = extension.

make k the subject of the equation,

k = 2E/e².................. Equation 2

Given: E = 35 J, e = 0.54 m

Substitute into equation 2

k = 2(35)/0.54²

k = 70/0.2916

k = 240.1 N/m.

Hence the spring constant of the spring is 240.1 N/m