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

Correct Option - 3


Detailed Solution:

$\Rightarrow \mathrm{T}=\mathrm{W}+\mathrm{f}$

$=20000+3000$

$=23000\mathrm{N}$

$\Rightarrow$ Power $=\mathrm{T}\mathrm{v}$

$=23000\times 1.5$

$=34500$ watts 

<p><span contenteditable="false"> <math> <mo>⇒</mo> <mi mathvariant="normal">T</mi> <mo>=</mo> <mi mathvariant="normal">W</mi> <mo>+</mo> <mi mathvariant="normal">f</mi> </math> </span></p><p><span contenteditable="false"> <math> <mo>=</mo> <mn>20000</mn> <mo>+</mo> <mn>3000</mn> </math> </span></p><p><span contenteditable="false"> <math> <mo>=</mo> <mn>23000</mn> <mtext> </mtext> <mi mathvariant="normal">N</mi> </math> </span></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1756963517phpwgrHE4.jpeg" alt="" width="78" height="82" loading="lazy"></picture></div></div><p><span contenteditable="false"> <math> <mo>⇒</mo> </math> </span>Power <span contenteditable="false"> <math> <mo>=</mo> <mi mathvariant="normal">T</mi> <mi mathvariant="normal">v</mi> </math> </span></p><p><span contenteditable="false"> <math> <mo>=</mo> <mn>23000</mn> <mo>×</mo> <mn>1.5</mn> </math> </span></p><p><span contenteditable="false"> <math> <mo>=</mo> <mn>34500</mn> </math> </span> watts&nbsp;</p>

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