Work Energy Power

JEE Main level, numerical answer, medium

93 of 183

Question 93 of 183

An engine is attached to a wagon through a shock absorber of length 1.5m.1.5m. The system with a total mass of 40,000kg40,000kg is moving with a speed of 72kmh−172km{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 1.0m1.0m. If 9090% of energy of the wagon is lost due to friction, the spring constant is _________\times {10}^{5}N{m}^{-1}.

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Answer: 16

Work =ΔK.E.=\Delta K.E.
Wtriction+Wspring=0−12mv2{W}_{\text{triction}}+{W}_{\text{spring}}=0-\dfrac{1}{2}m{v}^{2}
−90100(12mv2)+Wspring=−12mv2-\dfrac{90}{100}\left(\dfrac{1}{2}m{v}^{2}\right)+{W}_{\text{spring}}=-\dfrac{1}{2}m{v}^{2}
wspring=−10100×12mv2{w}_{\text{spring}}=-\dfrac{10}{100}\times \dfrac{1}{2}m{v}^{2}
−12kx2=−120mv2⇒k=40000×(20)210×(1)2=16×105-\dfrac{1}{2}k{x}^{2}=-\dfrac{1}{20}m{v}^{2}\Rightarrow k=\dfrac{40000\times (20{)}^{2}}{10\times (1{)}^{2}}=16\times {10}^{5}

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