A person pushes a box on a rough horizontal platform surface. He applies a force of 200 <math> <mi mathvariant="normal">N</mi> </math> over a distance of <math> <mn>15</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math> . Thereafter, he gets progressively tired and his applied force reduces linearly with distance to <math> <mn>100</mn> <mtext> </mtext> <mi mathvariant="normal">N</mi> </math> . The total distance through which the box has been moved is <math> <mn>30</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math> . What is the work done by the person during the total movement of the box?

Sol. (n) λ=5 (n, m) : Integers2m+12λ=32⇒3/25=2m+12n⇒3n=10m+5N, m are integers.So, m=1, n=5, λ=1m=4n=15λ=13m=7n=25λ=15

A person pushes a box on a rough horizontal platform surface. He applies a force of 200 <math> <mi mathvariant="normal">N</mi> </math> over a distance of <math> <mn>15</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math> . Thereafter, he gets progressively tired and his applied force reduces linearly with distance to <math> <mn>100</mn> <mtext> </mtext> <mi mathvariant="normal">N</mi> </math> . The total distance through which the box has been moved is <math> <mn>30</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math> . What is the work done by the person during the total movement of the box? &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

<math> <mn>2780</mn> <mtext> </mtext> <mi mathvariant="normal">J</mi> </math> (Work and Energy)

Ask & Answer Question

Option 1 -

$5250 J$

Option 2 -

$5690 J$

Option 3 -

$3280 J$

Option 4 -

$2780 J$ (Work and Energy)

2 Views | Posted 4 months ago

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

Correct Option - 2

Detailed Solution:

Sol. (n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$