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The staples inside a stapler are kept in place by a spring with a relaxed length of 0.116 m. if the spring constant is 46.0 n/m, how much elastic potential energy is stored in the spring when its length is 0.146 m?

Respuesta :

Answer:

U = 0.0207 J

Explanation:

We know that:

U = [tex]\frac{1}{2}Kx^2[/tex]

where U is the potential energy, K the constant of the spring and x is the deformation.

so, the deformation is calcualted as:

x = 0.146m-0.116m

x = 0.03m

Finally, replacing the values of x and K, we get:

U = [tex]\frac{1}{2}(46n/m)(0.03m)^2[/tex]

U = 0.0207 J

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