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Work Power and Energy question

2023 · 10 Apr · Shift 2 · Q61
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Work Power and Energy question

2023 · 10 Apr · Shift 2 · Q61

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
If the maximum load carried by an elevator is 1400 kg1400 \mathrm{~kg}1400 kg(600 kg600 \mathrm{~kg}600 kg- Passengers + 800 kg\mathrm{kg}kg- elevator), which is moving up with a uniform speed of 3 m s−13 \mathrm{~m} \mathrm{~s}^{-1}3 m s−1 and the frictional force acting on it is 2000 N2000 \mathrm{~N}2000 N, then the maximum power used by the motor is ‾\underline{\hspace{2cm}}​ kW(g=10 m/s2)\mathrm{kW}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)kW(g=10 m/s2)
Numerical answer
View written solutionFree

Correct answer: 48

  1. Given data

    • Total mass lifted: m=600+800=1400 kgm = 600 + 800 = 1400\ \text{kg}m=600+800=1400 kg
    • Speed of elevator: v=3 m s−1v = 3\ \text{m s}^{-1}v=3 m s−1
    • Frictional force: f=2000 Nf = 2000\ \text{N}f=2000 N
    • Acceleration due to gravity: g=10 m s−2g = 10\ \text{m s}^{-2}g=10 m s−2

  2. Forces opposing upward motion

    Since the elevator is moving upward with uniform speed, its acceleration is zero. Therefore, the motor must balance the total downward opposing force:

    • Weight of elevator + passengers: mg=1400×10=14000 Nmg = 1400 \times 10 = 14000\ \text{N}mg=1400×10=14000 N
    • Frictional force: f=2000 Nf = 2000\ \text{N}f=2000 N

    So total resisting force is F=14000+2000=16000 NF = 14000 + 2000 = 16000\ \text{N}F=14000+2000=16000 N

  3. Power used by the motor

    Power is given by P=FvP = FvP=Fv

    Therefore, P=16000×3=48000 WP = 16000 \times 3 = 48000\ \text{W}P=16000×3=48000 W

  4. Convert into kW

    48000 W=48 kW48000\ \text{W} = 48\ \text{kW}48000 W=48 kW

  5. Final answer

    48\boxed{48}48​

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