A spiral spring is compressed by 0.1cm. calculate the elastic potential energy in the spring if the stiffness of the spring is 100Nm^1

Respuesta :

Answer:

[tex]E=5\times 10^{-5}\ J[/tex]

Explanation:

Given that,

A spring is compressed by 0.1 cm or 0.001 m

The spring constant of the spring, k = 100 N/m

The elastic potential energy in the spring is given by :

[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 100\times 0.001^2\\\\E=5\times 10^{-5}\ J[/tex]

So, the elastic potential energy of the spring is equal to [tex]5\times 10^{-5}\ J[/tex].

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