Q 11-05-163JEE MainJEE Main 2020 (7 Jan, Shift 1)Easy
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: ($1$ HP $= 746$ W, $g = 10\ \text{m s}^{-2}$)
Answer: (B) $1.9\ \text{m s}^{-1}$
At constant speed the motor must balance weight plus friction:
$$F = mg + f = 2000\times10 + 4000 = 24000\ \text{N}$$
$$v = \frac PF = \frac{60\times746}{24000} = \frac{44760}{24000} \approx 1.9\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics