Question

Initially mass m is held such that spring is in relaxed condition. If the mass m is suddenly released, maximum elongation in spring will be

Answer 1

ans- 2mg/k

explanation-

We have,

mass=m

According to the law of conservation of energy for the spring system

so,

Initial total energy = final total energy

Potential energy = kinetic energy + spring elastic energy

Now if x be an elongation of the spring,

Then,

mgx = (1/2)mv×mv +(1/2)kx×kx

Now, the maximum elongation corresponds to the point where the velocity becomes zero and the spring is about to turn backward.

So, by putting v=0 in the above relation we get

mgx= (1/2)kx×kx

Thus, the maximum elongation would be

X = 2mg/k.

Answer 2

Answer:

refer this attachment dear

Explanation: