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An engine is attahed to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,000kg is moving with a speed of `36kmh^(-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 `1.0m`. If `90%` of energy of the wagon is lost due to friction, calculate the spring constant. |
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Answer» Here,`l=1.5m, m=50000kg=5xx10^(4)kg` `upsilon=36kmh^(-1)=(36xx1000)/(60xx60)ms^(-1)=10ms^(-1), x=1.0m,k=?` Total KE`=(1)/(2)m upsilon^(2)=(1)/(2)xx5xx10^(4)(10)^(2)=2.5xx10^(6)J` As `90%` of energy of wagon is lost due to friction, therefore, energy transferred to shock absorber, `E=(10)/(100)xx2.5xx10^(6)=2.5xx10^(5)J` From `E=(1)/(2)kx^(2)` `k=(2E)/(x^(2))=(2xx2.5xx10^(5))/(1^(2))=5xx10^(5)N//m` |
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