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Q.Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms^2. The force of friction per quintal is 0.5 N.

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Subjective· 3mImportance★★★★★
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Adding the force for acceleration, the component of gravity along the incline, and the friction force gives a total required force of about 4.4 × 10⁵ N.

Given: mass of train, m=2000m = 2000 quintal =2000×100=2×105 kg= 2000 \times 100 = 2\times10^5\ kg (1 quintal = 100 kg); incline of 1 in 50, so sin⁡θ=150=0.02\sin\theta = \dfrac{1}{50} = 0.02; acceleration a=2 m/s2a = 2\ m/s^2; friction =0.5 N= 0.5\ N per quintal × 2000\times\ 2000 quintal =1000 N= 1000\ N; take g=9.8 m/s2g = 9.8\ m/s^2.

To move the train up the incline with the given acceleration, the applied force FF must supply three things simultaneously: the force to accelerate the mass, the force to overcome the component of gravity pulling it back down the incline, and the force to overcome friction.

F=ma+mgsin⁡θ+fF = ma + mg\sin\theta + f

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