(A) To maintain a constant speed, the engine must exert a force equal to the frictional force acting on the train.The frictional force $F_f$ is given

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(A) To maintain a constant speed, the engine must exert a force equal to the frictional force acting on the train.The frictional force $F_f$ is given by $F_f = \mu N$, where $\mu$ is the coefficient of kinetic friction and $N$ is the normal force.Since the train is on a horizontal track, $N = mg$.Given: $m = 3 \times 10^6 \,kg$, $\mu = 0.05$, $g = 10 \,m \,s^{-2}$, and velocity $v = 50 \,m \,s^{-1}$.$F_f = 0.05 \times (3 \times 10^6 \,kg) \times (10 \,m \,s^{-2}) = 0.05 \times 3 \times 10^7 \,N = 1.5 \times 10^6 \,N$.The power $P$ required is given by $P = F_f \times v$.$P = (1.5 \times 10^6 \,N) \times (50 \,m \,s^{-1}) = 75 \times 10^6 \,W = 75 \,MW$.

Class 11 Physics Newton's Laws of Motion and Friction Kinetic Friction and Motion on Rough Horizontal Surface

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The power required for an engine to maintain a constant speed of $50 \,m \,s^{-1}$ for a train of mass $3 \times 10^6 \,kg$ on rough rails is (The coefficient of kinetic friction between the rails and wheels of the train is $0.05$ and acceleration due to gravity $= 10 \,m \,s^{-2}$).

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$75 \,MW$
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$40 \,MW$
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$75 \,kW$
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$65 \,MW$

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Newton's Laws of Motion and Friction

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Kinetic Friction and Motion on Rough Horizontal Surface

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The power required for an engine to maintain a constant speed of $50 \,m \,s^{-1}$ for a train of mass $3 \times 10^6 \,kg$ on rough rails is (The coefficient of kinetic friction between the rails and wheels of the train is $0.05$ and acceleration due to gravity $= 10 \,m \,s^{-2}$).

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$A$ body of weight $64 \ N$ is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are $0.8$ and $0.6$ respectively,then the acceleration of the body will be

$A$ block of mass $5\, kg$ is on a rough horizontal surface and is at rest. Now a force of $24\, N$ is imparted to it with negligible impulse. If the coefficient of kinetic friction is $0.4$ and $g = 9.8\, m/s^2$,then the acceleration of the block is ........ $m/s^2$.

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