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An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 40,000 kg is moving with a speed of $72 {kmh}^{- 1}$ when the breaks 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, spring constant is _____ $× 105 N / m$

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answer is 16.

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Detailed Solution

Given, Mass of engine-wagon system, m=40,000kg

Velocity, $V = 72 × 5 / 18 = 20 m / s$

$K . E = 12 m v 2 = 12 × (40, 000) × (20) 2 = 8000000 J$

As 90%K.E of the system lost in friction, only 10% is transferred to spring. $∴ \frac{1}{2} K x^{2} = \frac{10}{100} \times 8000000$

$\Rightarrow \frac{1}{2} \times K \times 1 \times 1 = 800000$
$\Rightarrow K = 16 \times 10^{5} N / m$

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