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When a 4.10-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.90 cm. If the 4.10-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?

Respuesta :

Answer:

Explanation:

Using Hooke's law,

F = -kx

Where,

F = force of the mass

k = spring constant/stiffness

x = length of the spring

Given:

F1 = m1*a

= 4.1 * 9.81

= 40.221 N

F2 = m2*a

= 1.5 * 9.81

= 14.715 N

x1 = 2.9 cm

F1/x1 = F2/x2

Therefore,

x2 = (14.715 * 2.9)/40.221

= 1.06 cm.

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