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  • An electric lift with a maximum load of <img alt="2000 \mathrm{~kg} (\text{lift} + \text{passengers})" src="/latex-image/?2000%20%5Cmathrm%7B%7Ekg%7D%20%28%5Ctext%7Blift%7D%20&plus;%20%5Ct

An electric lift with a maximum load of 2000 \mathrm{~kg} (\text{lift} + \text{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)
 

Option: 1

20000


Option: 2

34500


Option: 3

23500


Option: 4

23000


Answers (1)

best_answer

For the minimum power delivered by the motor to the lift,

the lift is moving up with a constant speed i.e net acceleration is zero
\begin{aligned} \text { External force (by motor) } &=\text { Frictional force + weight } \\ &=3000 \mathrm{~N}+(2000) \times 10 \\ & =23000\end{aligned}

$$ \begin{aligned} P_{\text {delivered }} &=F \mathrm{~V} \\ &=(23000)(1.5) \\ &=34500 \mathrm{~W} \end{aligned}

Posted by

Rakesh

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