The power required for an engine to maintain | vedclass

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Question

(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

Vedclass · Accepted Answer

$75 \,MW$

Vedclass · Answer

(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: $m = 3 	imes 10^6 \,kg$, $\mu = 0.05$, $g = 10 \,m \,s^{-2}$, and velocity $v = 50 \,m \,s^{-1}$.$F_f = 0.05 	imes (3 	imes 10^6 \,kg) 	imes (10 \,m \,s^{-2}) = 0.05 	imes 3 	imes 10^7 \,N = 1.5 	imes 10^6 \,N$.The power $P$ required is given by $P = F_f 	imes v$.$P = (1.5 	imes 10^6 \,N) 	imes (50 \,m \,s^{-1}) = 75 	imes 10^6 \,W = 75 \,MW$.

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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