Question

For the situation shown in the given figure, block $A$ of mass $10kg$ is at the ground. The block $B$ is released from rest when the light spring is unstretched. The minimum mass of the block $B$ in kg such that block $B$ may leave the contact of ground is ______.

Answer 1

Answer: 5

When the block $A$ just starts to leave the surface, then.
$k{x}_0=Mg$
For minimum value of $m,x_0$ should be maximum elongation in the spring. At the instant of maximum elongation, the velocity of block $B$ will be zero.
$\therefore$ Loss in gravitational potential energy of block $B=$ Gain in elastic potential energy by spring
$\therefore mg{x}_0=\frac{1}{2}k{x}_0^2$
$\therefore x_0=\frac{2mg}{k}$
or $\frac{Mg}{k}=\frac{2mg}{k}$
$\Rightarrow m=\frac{M}{2}=5kg$