MPSC State Service 2017 · GS · Question 52 of 100

An elevator carrying, a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms-1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator. (Take g = 10 m/s2)

Physics

Answer: A. 59 hp

Power = force × velocity = (1800×10 + 4000) × 2 = 44000 W ≈ 59 hp.

Source: MPSC Rajyaseva Prelims 2017 GS Paper I (official paper)

Practice the full MPSC State Service 2017 paper

Timed test with all 100 questions, just like the real exam.

MPSC State Service 2017 · GS · Question 52 of 100

Physics

Power = force × velocity = (1800×10 + 4000) × 2 = 44000 W ≈ 59 hp.

Source: MPSC Rajyaseva Prelims 2017 GS Paper I (official paper)

Practice the full MPSC State Service 2017 paper

Timed test with all 100 questions, just like the real exam.