Question

What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm from its relaxed length?

Answer 1

The spring constant of the spring is 8888.8 N/m

Step-by-step explanation:

Given that,

Elastic potential energy of the spring, U = 25 J

Compression in the spring, x = 7.5 cm = 0.075 m

We need to find the spring constant of the spring. The elastic potential energy of the spring is given by the below formula as :

`E=(1)/(2)kx^2`

k is the spring constant

`k=(2E)/(x^2)`

`k=(2* 25)/((0.075)^2)`

k = 8888.8 N/m

So, the spring constant of the spring is 8888.8 N/m. Hence, this is the required solution.