A 60 H P  electric motor lifts an elevator having a maximum total load capacity of  2000 k g . If the frictional force on the elevator is 4000 N  , the speed of the elevator at full load is close to: 1 H P = 746 W , g = 10 m s - 2

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>1.5</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>2.0</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>1.7</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>1.9</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mo>/</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> </span></p>
9 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
V
7 months ago
Correct Option - 4
Detailed Solution:

E ? = 9 s i n ? 1.6 * 10 3 x + 48 * 10 10 t k ˆ ? / m

4000 * V + m g * V = P

60 * 746 2000 * 4000 = V V = 1.86 m / s 1.9 m / s

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Physics Work, Energy and Power 2021

Physics Work, Energy and Power 2021

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