An engine is attached to a wagon through a shock absorber of length $1.5 \, m$. The system with a total mass of $50,000 \, kg$ is moving with a speed of $36 \, km \, h^{-1}$ 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 the energy of the wagon is lost due to friction,calculate the spring constant.

(C) Given mass $m = 50,000 \, kg$ and initial velocity $v = 36 \, km/h = 10 \, m/s$.
The initial kinetic energy $K$ is given by $K = \frac{1}{2} m v^2 = \frac{1}{2} \times 50,000 \times (10)^2 = 2.5 \times 10^6 \, J$.
Since $90 \%$ of the kinetic energy is lost due to friction,only $10 \%$ of the energy is stored in the spring of the shock absorber.
Energy stored in the spring $\Delta E = \frac{1}{2} k x^2$,where $x = 1.0 \, m$ is the compression.
According to the problem,$\Delta E = 10 \% \text{ of } K = 0.10 \times 2.5 \times 10^6 \, J = 2.5 \times 10^5 \, J$.
Equating the two expressions for energy: $\frac{1}{2} k (1.0)^2 = 2.5 \times 10^5 \, J$.
Solving for $k$: $k = 2 \times 2.5 \times 10^5 = 5.0 \times 10^5 \, N/m$.