Class 11th Laws of Motion Physics Physics Laws of Motion

A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is: Given m = 8 kg, M = 16kg. Assume all the surfaces shown in the figure to be frictionless.

Option 1 - <p> <span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>6</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mi>g</mi> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> <mi>g</mi> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>g</mi> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>g</mi> </mrow> </math> </span></p>

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Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9} g$

Now for 8 kg,

$8 g s i n 30 ° + 8 a c o s 30 ° = m a_{1}$

$a_{1} = g * \frac{1}{2} + \frac{\sqrt[]{3}}{9} g * \frac{\sqrt[]{3}}{2}$

$= \frac{2}{3} g$

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