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Law of Conservation of Mechanical Energy

In figure a s...

Question

In figure a string attached to $m_{1}$ passes over smooth pulleys $p_{1}$ and $p_{2}$ and is then attached to right support S.
If $m_{1} = 2 k g a n d m_{2} = 4 k g,$ find the acceleration with which $m_{2}$ descends.
Given : $g = 10 m s^{- 2} .$ The mass of pulley $p_{2}$ is negligible.

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Solution

From force balnace in $F B D (1), T = m_{1} a_{1}$ $e q (1)$

From force balance in $F B D (2), m_{2} g - T_{1} = m_{2} a_{2}$ $e q (2)$

From force balance of pulley $P_{2}, 2 T = T_{1}$ (since pulley is massless)

We know that work done by $T e n s i o n$ is $z e r o$
$⟹ p o w e r$ $s u p p l i e d$ $b y$ $T e n s i o n = z e r o$
Power by $T e n s i o n = \to T . \to V = T V cos θ$ wher is velocity

Let $_{1}$ be the velocity of $m_{1}$ and $_{2}$ be the velocity of $m_{2}$

$T V_{1} cos 0 + T_{1} V_{2} cos 180 = 0$ (since power supplied by $t e n s i o n = 0$ )
$⟹ T V_{1} - 2 T V_{2} = 0$ (since $T_{1} = 2 T$ )
$V_{1} = 2 V_{2}$
Similar relation will be for $a c c e l e r a t i o n s$
$⟹ a_{1} = 2 a_{2}$
solving $e q (1) a n d e q (2)$
$T = 2 a_{1}$ and $4 g - 2 T = 4 \frac{a_{1}}{2} = 2 a_{1}$

$a_{1} = 2 \frac{g}{3}$
$⟹ a_{2} = \frac{g}{3} = \frac{10}{3} m / s^{2}$
$a_{2}$ is the acceleration with which mass $m_{2}$ will descend

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