Question

An engine is attached to a wagon through a shock absorber of length $1.5m$. The system with a total mass of $40,000kg$ is moving with a speed of $72km{h}^{-1}$ 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 $1.0m$. If $90\%$ of energy of the wagon is lost due to friction, the spring constant is $......\times 10^{5}N/m$

Answer 1

Answer: 16

Work $=\Delta K.E$
$W_{\text{friction }}+W_{\text{spring }}=0-\frac{1}{2}m{v}^2$
$-\frac{90}{100}\left(\frac{1}{2}m{v}^2\right)+W_{\text{Spring }}=-\frac{1}{2}m{v}^2$
$W_{\text{spring }}=-\frac{10}{100}\times \frac{1}{2}m{v}^2$
$-\frac{1}{2}k{x}^2=-\frac{1}{20}m{v}^2$
$\Rightarrow k=\frac{40000\times (20{)}^2}{10\times (1{)}^2}$
$=16\times 10^5$