Aspring has a force constant of 559.9 n/m. find the potential energy stored in the spring when the spring is a) stretched 4.54 cm from equilibrium. answer in units of j.

The potential Energy of the spring = 0.577 J.

Explanation:

Potential Energy of a spring: This is the energy stored in a spring when it is stretched or compressed. The S.I unit of potential Energy is (J).

Mathematically, the energy stored in a spring is given as,

E = 1/2ke² Equation 1.

Where E = potential Energy, k = spring constant, e = extension.

Given: k = 559.9 N/m, e = 4.54 cm = 4.54/100 = 0.0454 m.

Substituting into equation 1

E = 1/2(559.9)(0.0454)²

E = 0.577 J.

Thus The potential Energy of the spring = 0.577 J.

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