Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202310 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2023 · 10 April · Evening Shift · Physics · Work, Power and Energy

If the maximum load carried by an elevator is 1400kg (...

If the maximum load carried by an elevator is 1400kg1400kg (600kg-Passengers+800kg-elevator)( 600kg\text{-Passengers} + 800kg\text{-elevator} ) , which is moving up with a uniform speed of 3ms−13m{s}^{- 1} and the frictional force acting on it is 2000N2000N, then the maximum power used by the motor is _____________kW.(g=10ms−2)kW . \left(g = 10m{s}^{- 2}\right)

If the maximum load carried by an elevator is 1400kg ( 600kg-Passengers + 800kg-elevator ) , which is moving up with a

Correct answer

0

Step-by-step solution

If the maximum load carried by an elevator is 1400kg ( 600kg-Passengers + 800kg-elevator ) , which is moving up with a

The maximum force acting is, {F}_\max = \left(14000 + 2000\right)N

=16000N= 16000N

The formula for power is given by, P=F\maxvP = {F}_\maxv.

It is given, v=3ms−1v = 3m{s}^{- 1}.

⇒P=16000N×3ms−1=48000W=48kW⇒ P = 16000N \times 3m{s}^{- 1} = 48000W = 48kW

⇒W=48kW⇒ W = 48kW

Practice Work, Power and Energy on Quantrex Academy →

More from Work, Power and Energy

All Work, Power and Energy questions Full Work, Power and Energy list All JEE Main PYQs