Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The coefficient of static friction, $\mu_s$ , between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. ( $g = 10 \text{ m/s}^2$ )

2.0 kg

4.0 kg

0.2 kg

0.4 kg

Step-by-Step Solution

System Analysis: Block A is on a horizontal surface, held in place by the static friction. Block B is hanging and provides the tension in the string trying to pull Block A. For the blocks not to move, the system must be in equilibrium.
Forces on Block B: The tension $T$ in the string supports the weight of Block B. $T = m_B g$
Forces on Block A: The tension $T$ pulls Block A horizontally. This is opposed by the static friction force $f_s$ . The normal reaction on A is $N = m_A g$ . For equilibrium: $T = f_s$
Limiting Condition: The maximum mass of B corresponds to the maximum tension Block A can withstand without sliding. This occurs when static friction reaches its limiting value: $(f_s)_{max} = \mu_s N = \mu_s m_A g$ [Source 56, 63].
Calculation: Equating tension to limiting friction: $m_B g = \mu_s m_A g$ $m_B = \mu_s m_A$ $m_B = 0.2 \times 2 \text{ kg} = 0.4 \text{ kg}$

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