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  • A engines 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 when the brakes are applied to bring it to rest.

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 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 the energy of the wagon is lost due to friction, calculate the spring constant.

Answers (1)

Kinetic energy is equal to

\frac{1}{2} mv^{2}

m = 50,000 kg, v = 10m/s

so, KE = \frac{1}{2} \times 50000 \times 10\times 10 = 2500000 J

now, as we know that 90% of the kinetic energy is lost as a result of friction brakes. Hence, only 10% of KE is transferred to spring.

So, the kinetic energy of spring would be

\frac{1}{2} kx^{2}

Here, x = 1m

Hence, K = 500000 J

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infoexpert24

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