Initially spring in its natural length. Now, a block of mass 0.25 kg is released, then find out the value of
maximum force exerted by system on the floor. [AIIMS 2019]

If x be the compression in the spring, when a block of 0.25 kg is released.
From law of conservation of energy,
Potential energy of spring = Potential energy of block

$\frac{1}{2}k{x}^2=mgx$

where, k is force constant of spring,

$kx=2mg=2\times 0.25g$

$kx=0.5 g$

$\therefore$  Force applied by the spring on the block,  $F=kx=0.5 \mathrm{g}$

From free body diagram.

$\therefore$ Maximum force by system on the floor,

$N=F+2 g=0.5 g+2 g=2.5 g$

$\left(\because g=10{ \mathrm{ms}}^{-2}\right)$

             = 25N