Q103 / 180 · JIPMER 2018 · Physics
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 $\mathrm{v}=0.5 \mathrm{~m} / \mathrm{s}$
- A$5 kW$
- B$1.5 kW$
- Correct answerC$5.5 kW$
- D$4.5 kW$
Solution
$\because$ Net force required to lift a hook and load,
$\mathrm{F}_{\text {net }}=1000+10000=11000 \mathrm{~N}$
$\because$ Power required to lift hook, $\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}$
$\begin{aligned} \because \quad \mathrm{W} & =\mathrm{F}_{\text {net }} \mathrm{d} \\ \mathrm{P} \times \mathrm{t} & =\mathrm{F}_{\text {net }} \mathrm{d} \\ \mathrm{P} & =\frac{\mathrm{F}_{\text {net }} \mathrm{d}}{\mathrm{t}}=\mathrm{F}_{\text {net }}\left(\frac{\mathrm{d}}{\mathrm{t}}\right)=\mathrm{F}_{\text {net }} \mathrm{V} \quad\left[\text { Here, } \mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}\right]\end{aligned}$
or $\mathrm{P}=11000 \times 0.5=5500 \mathrm{~W}=5.5 \mathrm{~kW}$
Hence, the correct answer is option: 'C'.