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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.
Solution
$m=50,000 \mathrm{~kg}$
$v=36 \mathrm{~km} / \mathrm{h}$
$=\frac{36 \times 1000}{60 \times 60}$
$=10 \mathrm{~m} / \mathrm{s}$
$x =1.0 \mathrm{~m}$
$\mathrm{~K} =\frac{1}{2} m v^{2}$
$=\frac{1}{2} \times 50000 \times(10)^{2}$
$=25000 \times 100 \mathrm{~J}$
$=2.5 \times 10^{6} \mathrm{~J}$
Since, $90 \%$ of $\mathrm{KE}$ of the system is lost due to friction, therefore $10 \%$ energy transferred to shock absorber, is given by
$\Delta \mathrm{E}=\frac{1}{2} k x^{2}$
$\Delta \mathrm{E}=10 \% \text { of } \mathrm{K}$
$\therefore \frac{1}{2} k x^{2}=\frac{10}{100} \times 2.5 \times 10^{6} \mathrm{~J}$
$\therefore k=5.0 \times 10^{5} \mathrm{~N} / \mathrm{m}$
