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

Question

10.1.1. 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 object of mass 2m is attached to the bottom of the second spring, how far will the bottom of the second spring move downward from its outstretched position? Assume the masses of the springs are negligible when compared to m.

Explanation / Answer

here,

the mass is m and the distance is y

let the spring constant be k

for the spring in equilibrium

k * y = m*g

k = mg/y ....(1)

when 2m mass is hung

when the seccond springs was attached

let the spring constant be K' = K/2

then

as spring is in equilibrium

equating the forces

k' * x = 2 * m * g

x * m*g /(2*y) = 2 * m * g

x = 4 y

the bottom of the second spring move downward from its outstretched position is 4y

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