JEE Main 2020 · 7 January · morning · Single correct · +4 / −1

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 ms⁻²)

  1. (a) 1.5 ms⁻¹
  2. (b) 1.7 ms⁻¹
  3. (c) 2.0 ms⁻¹
  4. (d) 1.9 ms⁻¹
Answer and solution

Answer (d) 1.9 ms⁻¹

Solution

Physics sir · 3 steps

At steady speed the motor pushes against weight plus friction, so divide its power by that total force.

  1. Motor power. Convert horsepower to watts: P = 60 × 746 = 44760 W.
  2. Force to overcome. At full load and constant speed, the motor must balance the weight mg = 2000 × 10 = 20000 N and the friction 4000 N. Friction opposes the upward motion, so it adds to the weight: F = 24000 N.

    v = P/F = (44760 W)/(24000 N) ≈ 1.87 m s⁻¹

  3. So. The speed is about 1.87 m s⁻¹, which is closest to 1.9 m s⁻¹, option (d).

The idea to keep. For anything moving at constant speed, P = Fv, where F is the total resisting force. When lifting, add the weight and the friction before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Work, Energy & Power

Asked about 100 times in JEE Main since 2020. This question is from 2020.