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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m.What is the displacement of the spring ?

Respuesta :

dexteright02
Data:
[tex]E_{pe} = 5184J[/tex]
[tex]K(constant) = 16200N/m[/tex]
[tex]x(displacement) = ?[/tex]


For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

[tex] E_{pe} = \frac{k*x^2}{2} [/tex]


Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

[tex] E_{pe} = \frac{k*x^2}{2} [/tex]
[tex] 5184 = \frac{16200*x^2}{2} [/tex]
[tex]5184*2 = 16200*x^2[/tex]
[tex]10368 = 16200x^2[/tex]
[tex]16200x^2 = 10368[/tex]
[tex] x^{2} = \frac{10368}{16200} [/tex]
[tex] x^{2} = 0.64[/tex]
[tex]x = \sqrt{0.64} [/tex]
[tex]\boxed{\boxed{x = 0.8m}}[/tex]

Answer:

The displacement of the spring = 0.8m

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