Question

A $60HP$ electric motor lifts an elevator having a maximum total load capacity of $2000kg$. If the frictional force on the elevator is $4000N$, the speed of the elevator at full load is close to $(1HP=7464W,G=10m/s^2)$

• A. $1.7m/s$
• B. $2.0m/s$
• C. $1.9m/s$
• D. $1.5m/s$

Answer 1

The correct option is C $1.9m/s$

Step1: Draw a free body diagram of the given problem.

Step2: Find the tension in the cable.

Given, power $p=60HP,$

Mass of lift $m=2000kg,$

frictional force on the elevator $=4000N$

Let us assume elevator is moving upward with constant speed $v$.From $F.B.D$ of elevator Tension in cable,

$T=2000g+f_r$

$=20000+4000$

$T=24000N$

Step3: Find the power supplied by the motor.

Formula Used: $p=F.v$

The power supplied by motor,
Power $P$ = Tension in cable $x$ speed

$=T\times v$

$60\times 746=\left(24000\right)v$

$v=\frac{60\times 746}{24000}$

$=1.865\approx 1.9m/s$

Final Answer: (c)