Question

Two discs, each having mass m, are attached rigidly to the ends of a vertical spring. One of the discs rests on a horizontal surface and the other produces a compression $x_0$ on the spring when it is in equilibrium. How much further must the spring be compressed so that when the force causing compression is removed, the extension of the spring will be able to lift the lower disc of the table?

• A. $x_0$
• B. $2x_0$
• C. $3x_0$
• D. $1.5x_0$

Answer 1

Answer: (B)

$2x_0$

Let $x_0$ be the equilibrium compression in the spring due to weight of the disc.

$⟹x_0=\frac{mg}{k}.........\left(1\right)$

Now if spring is further compressed by $x$, it rebounds to $P^{\prime }$.

Due this the actual extension in the spring is $x-x_0$

So this $k(x-x_0)$ spring force just lifts the lower disc

$⟹k(x-x_0)=mg⟹x=\frac{mg}{k}+x_0$

Using (1)

$x=2x_0$

hence Option B is correct