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

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Answer & Solution

Correct Answer

(A) \(16\)

Detailed explanation

\(\text { Work }=\Delta K . E\)

\(W_{\text {friction }}+W_{\text {spring }}=0-\frac{1}{2} m v^{2}\)

\(-\frac{90}{100}\left(\frac{1}{2} m v^{2}\right)+W_{\text {Spring }}=-\frac{1}{2} m v^{2}\)

\({W}_{\text {spring }}=-\frac{10}{100} \times \frac{1}{2} {mv}^{2}\)

\(-\frac{1}{2} {kx}^{2}=-\frac{1}{20} {mv}^{2}\)

\(\Rightarrow {k}=\frac{40000 \times(20)^{2}}{10 \times(1)^{2}}=16 \times 10^{5}\)

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