question_answer 71)

                  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 energy of the wagon is lost due to friction, calculate the spring constant.                

Answer:

                  \[m=5\times {{10}^{4}}\,\,kg,\,\,\upsilon =36\text{ }km\text{ }{{h}^{1}}\]                 \[=36\times \frac{5}{18}=10\,m\,{{s}^{-1}}\]                 K.E. of the system \[=\frac{1}{2}\,m{{\upsilon }^{2}}\]                 \[=\frac{1}{2}\,\times \,5\times \,{{10}^{4}}\,\times 100\,=2.5\,\,\times {{10}^{6}}\,J\]                  Since 90% of the kinetic energy of the wagon is lost due to friction, so only 10% of the kinetic energy is stored as the elastic potential of the shock absorber. That is,                 \[\frac{1}{2}\,k{{x}^{2}}=\,10%\] of \[K.E.\,=\,\frac{10}{100}\,\times 2.5\,\,\times {{10}^{6}}\]                 \[=2.5\times {{10}^{5}}\]                         or \[k=\,\frac{5.0\,\times \,{{10}^{-5}}}{{{x}^{2}}}\,=\,\frac{5.0\,\times {{10}^{5}}}{1}\] \[\left( \because \,\,k=1m \right)\]                 \[\therefore \] \[k=5.0\,\,\times \,{{10}^{5}}\,N{{m}^{-1}}\]