Q 11-05-203NEETNEET 2022Top questionEasy
An electric lift with a maximum load of $2000$ kg (lift + passengers) is moving up with a constant speed of $1.5\ \text{m s}^{-1}$. The frictional force opposing the motion is $3000$ N. The minimum power delivered by the motor to the lift in watts is : ($g = 10\ \text{m s}^{-2}$)
Answer: (D) $34500$
At constant speed the motor's pull balances weight plus friction:
$$F = mg + f = 2000 \times 10 + 3000 = 23000\ \text{N}$$
$$P = Fv = 23000 \times 1.5 = 34500\ \text{W}$$
Solution by Sreeraj P, M.Sc Physics