Question

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 $...... \times 10^{5}\, N / m$

Answer 1

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