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: $\left( 1 \mathrm { HP } = 746 \mathrm {~W} , g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 } \right)$\\
(1) $1.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(2) $1.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(3) $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(4) $2.0 \mathrm {~m} \mathrm {~s} ^ { - 1 }$