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In the system shown in Figure The masses of the bodies are known to be $m_{1}$ and $m_{2}$ , the coefficient of friction between the body $m_{1}$ and the horizontal plane is equal to $μ$ , and a pulley of mass $m$ is assumed to be a uniform disc. The thread does not slip over the pulley. At the moment $t = 0$ the body $m_{2}$ starts descending. Assuming the mass of the thread and the friction in the axle of the pulley to be negligible, find the work performed by the friction forces acting on the body $m_{1}$ over the first $t$ second after the beginning of motion.

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$\frac{μ m_{1} g^{2} t^{2} (2 m_{2} - μ m_{1})}{(2 m_{1} + 2 m_{2} + m)}$

$\frac{μ m_{1} g^{2} t^{2} (m_{2} - μ m_{1})}{(2 m_{1} + 2 m_{2} + m)}$

$\frac{μ m_{1} g^{2} t^{2} (m_{2} - μ m_{1})}{(2 m_{1} + 2 m_{2})}$

$2 \frac{μ m_{1} g^{2} t^{2} (m_{2} - μ m_{1})}{(2 m_{1} + 2 m_{2} + m)}$

answer is C.

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Detailed Solution

Let a is the acceleration of the blocks. Equation of motion of the blocks are

$T_{1} - μ N = m_{1} a \dots (1) m_{2} g - T_{2} = m_{2} a \dots \dots (2)$

and for the rotation of the pulley

$T_{2} R - T_{1} R = 1 α = I (a / R) \dots (3)$

where $I = \frac{m R^{2}}{2}$

Solving above equations, we get

$a = \frac{(m_{2} - μ m_{1}) g}{m_{1} + m_{2} + \frac{m}{2}}$

The distance travelled by the block in $t$ second

$s = \frac{1}{2} a t^{2}$

Work done by the friction

$W = f_{r} \times s = μ N \times \frac{1}{2} a t^{2} = \frac{(μ m_{1} g) \times \frac{1}{2} (m_{2} - μ m_{1}) g}{(m_{1} + m_{2} + \frac{m}{2})} t^{2} = \frac{μ m_{1} g^{2} t^{2} (m_{2} - μ m_{1})}{(2 m_{1} + 2 m_{2} + m)}$

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