Explanation

(d) Let $m$ be the mass of the block, $h$ the height from which it is dropped, and $x$ the compression o the spring. Since, energy is conserved, so
Final gravitational potential energy
= final spring potential energy
or $mg(h + x) = \frac{1}{2}kx^{2}$
or $mg(h + x) + \frac{1}{2}kx^{2} = 0$
or $kx^{2} - 2mg(h + x) = 0$
$kx^{2} - 2mgx - 2mgh = 0$
This is a quadratic equation for $x$. Its solution is
$x = \frac{mg \pm \ \sqrt{(mg)^{2} + 2mghk}}{k}$
Now, $mg = 2 \times 9.8 = 19.6$ N
and $hk = 0.40 \times 1960 = 784$ N
$\therefore\ \ \ \ x = \frac{19.6 \pm \sqrt{(19.6)^{2} + 2(19.6)(784)}}{1960}$
$= 0.10$ m or $- 0.080$ m
Since, $x$ must be positive (a compression) we accept the positive solution and reject the negative solution. Hence, $x = 0.10$ m