Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A system consists of three masses $m_1$ , $m_2$ and $m_3$ connected by a string passing over a pulley P. The mass $m_1$ hangs freely and $m_2$ and $m_3$ are on a rough horizontal table (the coefficient of friction = $\mu$ ). The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_1$ is: (Assume $m_1=m_2=m_3=m$ )

$\frac{g(1-g\mu)}{9}$

$\frac{2g\mu}{3}$

$\frac{g(1-2\mu)}{3}$

$\frac{g(1-2\mu)}{2}$

Step-by-Step Solution

System Analysis: The three blocks move together with a common acceleration $a$ . The hanging mass $m_1$ pulls the system downwards, while the friction on masses $m_2$ and $m_3$ opposes the motion.
Forces:

Driving Force: Gravitational force on the hanging mass $m_1$ : $F_{gravity} = m_1g$ .
Resisting Force: Frictional force on the horizontal masses $m_2$ and $m_3$ . Since they are on a rough table with coefficient $\mu$ , the kinetic friction is $f_k = \mu N$ . Here $N = mg$ . Total Friction $f = f_2 + f_3 = \mu m_2g + \mu m_3g = \mu g(m_2 + m_3)$ .

Equation of Motion (Newton's Second Law): $F_{net} = M_{total} \times a$ $m_1g - \mu g(m_2 + m_3) = (m_1 + m_2 + m_3)a$
Substitution: Given $m_1 = m_2 = m_3 = m$ . $mg - \mu g(m + m) = (m + m + m)a$ $mg - 2\mu mg = 3ma$ $mg(1 - 2\mu) = 3ma$
Solving for acceleration ( $a$ ): $a = \frac{g(1 - 2\mu)}{3}$

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