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प्रश्न
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.
उत्तर
Given, mass of the system (m) = 50,000 kg
Speed of the system (v) = 36 km/h
= `(36 xx 1000)/(60 xx 60)` = 10 m/s
Compression of the spring (x) = 1.0 m
KE of the system = `1/2 mv^2`
= `1/2 xx 50000 xx (10)^2`
= 25000 × 100 J
= 2.5 × 106 J
Since 90% of KE of the system is lost due to friction, therefore, the energy transferred to the shock absorber is given by ΔE = `1/2 kx^2`
= 10% of total KE of the system
= `10/100 xx 2.5 xx 10^6 J` or k = `(2 xx 2.5 xx 10^6)/(10 xx (1)^2)`
= 5.0 × 105 N/m
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