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Standard XIII

Physics

Problem Solving

A fixed wedge...

Question

A fixed wedge with both surfaces inclined at $45^{\circ}$ to the horizontal is shown in the figure. A block $P$ of mass $m$ is held on the smooth plane by a light string which passes over a smooth pulley A and is attached to a block $Q$ of mass $3 m$ which rests on the rough plane. The system is released from rest. Given that the acceleration of each block is of magnitude $\frac{g}{5 \sqrt{2}}$ then, the tension in the string is

m g

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\frac{6 m g}{5 \sqrt{2}}

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\frac{m g}{2}

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\frac{m g}{4}

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Solution

The correct option is B $\frac{6 m g}{5 \sqrt{2}}$
Applying Newton's second law for mass $m$
$T - m g sin 45^{\circ} = m a$
$T - \frac{m g}{\sqrt{2}} = \frac{m g}{5 \sqrt{2}}$
Given, $a = \frac{g}{5 \sqrt{2}}$
$T = \frac{6}{5} \frac{m g}{\sqrt{2}}$

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Similar questions

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The correct option is B 6mg5√2Applying Newton's second law for mass m T−mgsin45∘=ma T−mg√2=mg5√2 Given, a=g5√2 T=65mg√2

The correct option is B $\frac{6 m g}{5 \sqrt{2}}$
Applying Newton's second law for mass $m$
$T - m g sin 45^{\circ} = m a$
$T - \frac{m g}{\sqrt{2}} = \frac{m g}{5 \sqrt{2}}$
Given, $a = \frac{g}{5 \sqrt{2}}$
$T = \frac{6}{5} \frac{m g}{\sqrt{2}}$