An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of $72 kmh^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, the spring constant is _ _ _ _ $\times 10^5 Nm^{-1}$.
Solution
10 % of K.E. = $\frac {1}{2} kx^2 $
$\therefore 0.1 \times \frac {1}{2} mv^2 = \frac {1}{2} kx^2 $
$\therefore 0.1 \times 40,000 \times (72 \times \frac {5}{18})^2 = k \times 1.0^2 $
$\Rightarrow k = 4,000 \times 20^2 = 16 \times 10^5 Nm^{-1} $
Answer: 16