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Laws of Motion

JEE Main level, single correct, medium

173 of 210

Question 173 of 210

A $60HP$ electric motor lifts an elevator having a maximum total load capacity of $2000kg.$ If the frictional force on the elevator is $4000N,$ the speed of the elevator at full load is close to : $\left(1HP=746W,g=10m{s}^{-2}\right)$

Show answer and solution

Answer: (B) $1.9m{s}^{-1}$

4000\times V+mg\times V=P

\dfrac{60\times 746}{4000+20000}=V

V=

1.9m{s}^{-1}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Laws of Motion (Frictional force), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.