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Amy pulls a spring with a spring constant k = 100 stretching it from its rest length of 0.10 m to 0.2

Respuesta :

Answer:

Elastic potential energy  E = 0.5 J

Explanation:

Let us assume we have to find the elastic potential energy stored in the spring.

Given: k = 100

change in the length of the spring x = 0.2-0.1 = 0.1 m

Elastic potential energy

E = [tex]\frac{1}{2} kx^2[/tex]

plugging the values we get

[tex]E = 0.5\times100\times0.1^2\\=0.5 J[/tex]

Hence, Elastic potential energy  E = 0.5 J

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