Question 152 of 183

If the maximum load carried by an elevator is $1400kg$ $(600kg\text{-Passengers}+800kg\text{-elevator})$ , which is moving up with a uniform speed of $3m{s}^{-1}$ and the frictional force acting on it is $2000N$, then the maximum power used by the motor is _____________$kW.\left(g=10m{s}^{-2}\right)$

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Answer: 48

The maximum force acting is, ${F}_{max}=\left(14000+2000\right)N$
$=16000N$
The formula for power is given by, $P={F}_{max}v$.
It is given, $v=3m{s}^{-1}$.
$\Rightarrow P=16000N\times 3m{s}^{-1}=48000W=48kW$
$\Rightarrow W=48kW$