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 $m{s}^{-1}$ ,    [JIPMER 2018]

Net force required to lift a hook and load,

$F_{\text{net }}=1000+10000=11000 \mathrm{N}$

Power required to lift the hook,  $P=\frac{W}{t}$

$As, W=F_{net} d$

$\therefore P=\frac{F_{\text{net }}d}{t}=F_{\text{net }}\left(\frac{d}{t}\right)=F_{\text{net }}v \left(\because v=\frac{d}{t}\right)$

or     $P=11000\times 0.5=5500 \mathrm{W}=5.5 \mathrm{kW}$