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 \(72\ \mathrm{kmh^{-1}}\) 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.
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 \(72\ \mathrm{kmh^{-1}}\) 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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Only 10% of the initial kinetic energy goes into the spring: \(0.1\times\tfrac12 mv^2 = \tfrac12kx^2\).
Updated On: Nov 4, 2026
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Correct Answer: 16
Solution and Explanation
\(v = 72\) km/h \(= 20\) m/s. Initial KE \(= \tfrac12(40000)(20)^2 = 8\times10^{6}\) J.
90% is lost to friction, so energy stored in the spring \(= 0.10\times8\times10^6 = 8\times10^{5}\) J.
\(\tfrac12kx^2 = 8\times10^5\) with \(x = 1.0\) m gives \(k = 1.6\times10^{6}\ \mathrm{N/m} = 16\times10^{5}\ \mathrm{N/m}\).
Answer: 16
90% is lost to friction, so energy stored in the spring \(= 0.10\times8\times10^6 = 8\times10^{5}\) J.
\(\tfrac12kx^2 = 8\times10^5\) with \(x = 1.0\) m gives \(k = 1.6\times10^{6}\ \mathrm{N/m} = 16\times10^{5}\ \mathrm{N/m}\).
Answer: 16
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