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 : ($1 \,HP = 746 \,W, \,g = 10\, ms^{-2}$ )

• A. $1.5 \,ms^{- 1}$
• B. $1.7 \,ms^{- 1}$
• C. $2.0 \,ms^{- 1}$
• D. $1.9 \,ms^{- 1}$

Answer 1

Answer: (D)

$1.9 \,ms^{- 1}$

Answer 2

$4000 ? V + mg ? V = P$
$\frac{60\times746}{4000+20000}=V$
$V=1.86\,m/s. \approx 1.9\,m/s.$