An ideal spring with spring constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially in an unstretched condition. Then, the maximum extension in the block from its mean position will be,

Question

An ideal spring with spring constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially in an unstretched condition. Then, the maximum extension in the block from its mean position will be, (A) 4(M g/K) (B) 2(M g/K) (C) (M g/K) (D) (M g/2 K)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-t0ukm53ficgwd2k6.jpg" /> Let x be the maximum extension of the spring. From energy conservation, loss in gravitational potential energy = gain in potential energy of spring. Mgx=(1/2)Kx2 ⇒ x=(2 M g/K) .

Q. An ideal spring with spring constant $K$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially in an unstretched condition. Then, the maximum extension in the block from its mean position will be,

$4 \frac{M g}{K}$

$2 \frac{M g}{K}$

$\frac{M g}{K}$

$\frac{M g}{2 K}$

Let $x$ be the maximum extension of the spring.
From energy conservation,
loss in gravitational potential energy $=$ gain in potential energy of spring.
$M gx = \frac{1}{2} K x^{2}$
$\Rightarrow$ $x = \frac{2 M g}{K}$ .

Q. An ideal spring with spring constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially in an unstretched condition. Then, the maximum extension in the block from its mean position will be,

4KMg​

2KMg​

KMg​