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

• A. mg/k
• B. 2mg/k
• C. mg/2k
• D. mg/4k

Answer 1

The correct option is B 2mg/k

The correct option is B.

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,

$mgx=\left(\frac{1}{2}\right)m{v}^2+\left(\frac{1}{2}\right)k{x}^2$

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

$mgx=\left(\frac{1}{2}\right)k{x}^2$

Thus, the maximum elongation would be

$x=\frac{2mg}{k}$