Question

A spring has a spring constant of 120 N/m. How much energy is stored in the spring as it is stretched a distance of 0.20 meter?

Answer 1

2.4J

Step-by-step explanation:

Given parameters:

Spring constant = 120N/m

Extension = 0.2m

Unknown:

Amount of energy = ?

Solution:

The energy stored in this stretched spring is called the elastic potential energy.

It can be derived using the expression below:

Elastic Potential energy =
`(1)/(2)` ke²

k is the elastic constant

e is the extension

Insert the parameters;

Elastic potential energy =
`(1)/(2)` x 120 x 0.2² = 2.4J