A spring is hung vertically from a fixed support. When an object of mass m is at

Question

A spring is hung vertically from a fixed support. When an object of mass m is attached to the end of the spring, it stretches by a distance y. When an object of mass of 2m is hung from the spring, it stretches by a distance 2y. A second, identical spring is then attached to the free end of the first spring. If the mass 2m is attached to the bottom of the second spring, how far will the bottom of the second spring move downward from its unstretched position? Assume the masses of the springs are negligible when compared to m.

Explanation / Answer

let,

spring constant is k

if mass m is attached, then stretched distance is y

if mass 2m is attached, then stretched distance is 2y

===> restoring force, F1=K*(2y)

and again,

if mass 2m is attached, stretched distance is 2y

and

restoring force, F2=k*(2y)

Fnet =F1+F2

Fnet= K(2y) + k(2y)

Fnet=k(4y)

here,

total elongation is 4y ---->

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