Question

What is the elastic potential energy of a spring that is compressed a distance of 0.35 m and has a spring constant of 71.8 N/m?

Answer 1

P.E = 4.398 Joules.

Step-by-step explanation:

Given the following data;

Spring constant, k = 71.8N/m

Displacement, x = 0.35m

To find the elastic potential energy;

The elastic potential energy of an object is given by the formula;

`P.E = \frac {1}{2}kx^(2)`

Substituting into the equation, we have;

`P.E = \frac {1}{2}*71.8 *(0.35)^(2)`

`P.E = 35.9 * 0.1225`

Elastic potential energy = 4.398 Joules.

Therefore, the elastic potential energy of the spring is 4.398 Joules.