PYQ NEET- Work Energy And Power L-4

Question: An electric lift with a maximum load of $2000 \mathrm{~kg}$ (lift + passengers) is moving up with a constant speed of $1.5 \mathrm{~ms}^{-1}$. The frictional force opposing the motion is $3000 \mathrm{~N}$. The minimum power delivered by the motor to the lift in watts is : $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$#

A) 23000

B) 20000

C) 34500

D) 23500

Answer: 34500#

Solution:#

Explanation $$ \begin{aligned} & F_{\text {up }}=2000 \mathrm{~g}+3000 \ & =23000 \mathrm{~N} \end{aligned} $$

Minimum power $\mathrm{P}{\min }=\vec{F} \cdot \vec{v}$ $$ \begin{aligned} & P{\min }=F v=23000 \times \frac{3}{2} \ & =34500 \mathrm{~W} \end{aligned} $$