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?

P.E = 4.398 Joules.

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;

Substituting into the equation, we have;

Elastic potential energy = 4.398 Joules.

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

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