A mass of 1 kg falls from a height of 1 m and lands on a mass less platform supported by a spring having spring constant 15 Nm^-1 as shown in the figure. The maximum compression of the spring is. (acceleration due to gravity = 10 ms^-2)

Let x be the maximum compression. Now, potential energy gain by spring = Potential energy loss by mass of 1kg

$\frac{\text{1}}{\text{2}}{\text{kx}}^{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}mg}\left(\text{1+x}\right)$

$\Rightarrow \frac{\text{1}}{\text{2}}{\text{\hspace{0.17em}×\hspace{0.17em}15\hspace{0.17em}×\hspace{0.17em}x}}^{\text{2}}\text{\hspace{0.17em}=1\hspace{0.17em}×\hspace{0.17em}10\hspace{0.17em}×\hspace{0.17em}}\left(\text{1+x}\right)$

$\Rightarrow$15x² = 20 (1 + x)

$\Rightarrow$3x² = 4(1 + x)

$\Rightarrow$3x² – 4x – 4 = 0

$\Rightarrow$3x² – 6x + 2x – 4 = 0

$\Rightarrow$3x (x – 2) + 2 (x – 2) = 0

$\Rightarrow$x = - 2/3 or x = 2

- ve value of ‘x’ means spring moves upward, so it is impossible. So x = 2