A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg . If the fr

Question

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg . If the frictional force on the elevator is 4000 N , the speed of the elevator at full load is close to ( $1 \mathrm{HP}=746 \mathrm{~W}, g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )\

Professor PC Thomas Classes · Answer

(d) : Net force act on lift,\ $F_{	ext {net }}=m g+f$\ Here, $m g$ acts downward and $f$ also acts downward.\ $F_{	ext {net }}=(2000 	imes 10+4000)=24000 \mathrm{~N}$\ Power $=F_{	ext {net }} 	imes$ speed\ Speed $=\frac{	ext { Power }}{F_{	ext {net }}}=\frac{60 	imes 746}{24000}$\ Speed $=1.865 \mathrm{~m} \mathrm{~s}^{-1} \approx 1.9 \mathrm{~m} \mathrm{~s}^{-1}$\