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  • A block of mass 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 : <img alt="Given

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 \: \: \mathrm{m}=8 \mathrm{~kg}, M=16 \mathrm{~kg}
Assume all the surfaces shown in the figure to be frictionless.
Option: 1 \frac{3}{5}g
Option: 2 \frac{4}{3}g
Option: 3 \frac{6}{5}g
Option: 4 \frac{2}{3}g

Answers (1)

best_answer


a\rightarrow acceleration of Block wrt wedge
A\rightarrow Acceleration of Wedge
In frame of wedge

ma= MA\cos \theta +mg\sin \theta
a= A\cos \theta +g\sin \theta \rightarrow \left ( 1 \right )
mg\cos \theta = mA\sin \theta+N\rightarrow \left ( 2 \right )
N\sin \theta = MA
N\times \frac{1}{2}= 16\times A
N= 32 A\rightarrow \left ( 3 \right )
From eqn (2) & (3)
8g\cos \left ( 30^{\circ} \right )= 8A\times \frac{1}{2}+32A
8g\times \frac{\sqrt{3}}{2}= 4A+32A= 36A
4g\sqrt{3}= 36A
A= \frac{\sqrt{3}g}{9}
a= A\cos 30^{\circ}+g\sin 30^{\circ}
= \frac{\sqrt{3}g}{9}\times \frac{\sqrt{3}}{2}+g\times \frac{1}{2}
= \frac{3g}{18}+\frac{9g}{18}
\rightarrow a = \frac{12g}{18}= \frac{2g}{3}
The correct option is (4)

Posted by

vishal kumar

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