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A spring on a horizontal surface can be stretched and held 0.6 m from its equilibrium position with a force of 78 N. a. How much work is done in stretching the spring 5.5 m from its equilibrium​ position? b. How much work is done in compressing the spring 1.5 m from its equilibrium​ position?

Respuesta :

Answer:

Step-by-step explanation:

x = 0.6 m

F = 78 N

Let k be the spring constant.

F = k x

78 = 0.6 k

k = 130 N/m

(a)

x = 5.5 m

W = 0.5 kx²

W = 0.5 x 130 x 5.5 x 5.5

W = 7865 J

(B) x = 1.5 m

W = 0.5 kx²

W = 0.5 x 130 x 1.5 x 1.5

W = 146.25 J

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