Mass of smaller body $\mathrm{m}=1 \mathrm{kg}$

Smaller mass $(\mathrm{m}=1 \mathrm{kg})$ $executes \,S.H.M \,of$ 

angular frequency $\omega=25$ rad $/ \mathrm{s}$

Amplitude $x=1.6 \mathrm{cm}=1.6 \times 10^{-2}$

As we know,

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$

$\boldsymbol{\alpha}, \quad \frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$

or, $\quad \frac{1}{25}=\sqrt{\frac{1}{\mathrm{K}}}[\because \mathrm{m}=1 \mathrm{kg} ; \omega=25 \mathrm{rad} / \mathrm{s}]$

or, $\quad \mathrm{K}=625 \mathrm{Nm}^{-1}$

The maximum force exerted by the system on the floor

$=\mathrm{Mg}+\mathrm{Kx}+\mathrm{mg}$

$=4 \times 10+625 \times 1.6 \times 10^{-2}+1 \times 10$

$=40+10+10$

$=60 N$