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Ne4ueva [31]
3 years ago
6

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?

Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

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;

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.

<em>Therefore, the elastic potential energy of the spring is 4.398 Joules. </em>

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