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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Work, Power and Energy

JIPMER 2018 · Physics · Work, Power and Energy

1000 N force is required to lift a hook and 10000 N force is...

1000 N force is required to lift a hook and 10000 N force is requires to lift a load slowly. Find power required to lift hook with load with speed v=0.5 m/s\mathrm{v}=0.5 \mathrm{~m} / \mathrm{s}

Options

  1. A5kW5 kW
  2. B1.5kW1.5 kW
  3. C5.5kW5.5 kW
  4. D4.5kW4.5 kW

Correct answer

C. 5.5kW5.5 kW

Step-by-step solution

∵\because Net force required to lift a hook and load, Fnet =1000+10000=11000 N∵\mathrm{F}_{\text{net }}=1000+10000=11000 \mathrm{~N}\because Power required to lift hook, P=Wt\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}∵W=Fnet dP×t=Fnet dP=Fnet dt=Fnet (dt)=Fnet V[ Here, v=dt]\begin{aligned} \because \quad \mathrm{W} & =\mathrm{F}_{\text{net }} \mathrm{d} \\ \mathrm{P} \times \mathrm{t} & =\mathrm{F}_{\text{net }} \mathrm{d} \\ \mathrm{P} & =\frac{\mathrm{F}_{\text{net }} \mathrm{d}}{\mathrm{t}}=\mathrm{F}_{\text{net }}\left(\frac{\mathrm{d}}{\mathrm{t}}\right)=\mathrm{F}_{\text{net }} \mathrm{V} \quad\left[\text{ Here, } \mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}\right]\end{aligned}orP=11000×0.5=5500 W=5.5 kWor \mathrm{P}=11000 \times 0.5=5500 \mathrm{~W}=5.5 \mathrm{~kW}
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