The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is $16.0 N$ when the system is set into motion. Then the decrease in the gravitational potential energy during the first second after the system is released from rest is Joule .(take $g = 10 m / s^{2}$ )

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Solution

Work done by the tension on the system is zero i.e. $W_{T} |_{s y s t e m} = 0$
Now, work done by the gravity=change in kinetic energy,
$W_{g} = Δ K . E .$
Initially when the system is in rest then system have only gravitational potential energy. As the system start accelerating some part of this energy converted into kinetic energy and the remaining part will be the potential energy.
Increase in K.E.
K.E $= \frac{1}{2} m v^{2} + \frac{1}{2} 2 m v^{2}$ (velocity is same for both blocks)
$\begin{matrix} K . E . = \frac{3}{2} m v^{2} \\ (1) \end{matrix}$
and acceleration for Atwood machine,
$a = \frac{m_{2} - m_{1}}{m_{2} + m_{1}} g$
$a = \frac{2 m - m}{2 m + m} g$
$a = \frac{g}{3}$
From first equation of motion,
$v = u + a t$
$v = 0 + \frac{g}{3} \times 1 = \frac{g}{3}$
For Atwood machine Tension is given as,
$T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}}$
$16 = \frac{2 (2 m) (m) g}{3 m}$
$m = \frac{12}{g}$

Putting $m$ and $v$ in equation (1)
$K . E = \frac{3}{2} \times \frac{12}{9} \times \frac{g^{2}}{9} = 2 g = 20 J$

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