JEE Main Physics question
A 1000 kg lift is supported by a cable that can support 2000 kg. The shortest distance in which the lift can be stopped when it is descending with a speed of 2.5 ms⁻¹ is (take g = 10 ms⁻²)
Answer: (D)
Solution
Maximum available deceleration: − mg = ⟹ = (2000−1000)(10)/1000 = 10 ms⁻². s = v²/2a = 6.25/20 = 5/16 m.