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(C) Let the elevator move upward with a constant speed $V$.The total downward force acting on the elevator is the sum of the gravitational force and t

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$1.9$

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(C) Let the elevator move upward with a constant speed $V$.The total downward force acting on the elevator is the sum of the gravitational force and the frictional force.$T = mg + f_r$Given $m = 2000\; kg$,$g = 10\; ms^{-2}$,and $f_r = 4000\; N$.$T = (2000 	imes 10) + 4000 = 20000 + 4000 = 24000\; N$.The power $P$ of the motor is given by $P = T 	imes V$.Given $P = 60\; HP = 60 	imes 746\; W = 44760\; W$.Equating the power,we get:$44760 = 24000 	imes V$$V = \frac{44760}{24000} = 1.865\; m/s$.Rounding to one decimal place,the speed is approximately $1.9\; m/s$.

$A$ $60\; HP$ electric motor lifts an elevator having a maximum total load capacity of $2000\; kg$. If the frictional force on the elevator is $4000\; N$,the speed of the elevator at full load is close to .............. $m/s$.
$(1\; HP = 746\; W, g = 10\; ms^{-2})$

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$A$ $60\; HP$ electric motor lifts an elevator having a maximum total load capacity of $2000\; kg$. If the frictional force on the elevator is $4000\; N$,the speed of the elevator at full load is close to .............. $m/s$.$(1\; HP = 746\; W, g = 10\; ms^{-2})$