Three blocks A, B, and C of masses $4 ext{ kg}$, $2 ext{ kg}$, and $1 ext{ kg}$ respectively, are in con

Question

Three blocks A, B, and C of masses $4 	ext{ kg}$, $2 	ext{ kg}$, and $1 	ext{ kg}$ respectively, are in contact on a frictionless surface, as shown. If a force of $14 	ext{ N}$ is applied to the $4 	ext{ kg}$ block, then the contact force between A and B is:

Accepted Answer

1. **System Acceleration:** Since the blocks are in contact and moving together under a single force on a frictionless surface, they share a common acceleration $a$. According to Newton's Second Law ($F = ma$) [NCERT Class 11, Physics Part I, Sec 4.5]:
   $$ a = \frac{	ext{Total Force}}{	ext{Total Mass}} = \frac{14}{4 + 2 + 1} = \frac{14}{7} = 2 	ext{ m/s}^2 $$
2. **Contact Force Analysis:** To find the contact force between block A and block B ($F_{AB}$), we can analyze the forces acting on the combined system of blocks B and C. The force $F_{AB}$ is the external force responsible for accelerating masses B and C.
   $$ F_{AB} = (m_B + m_C) 	imes a $$
   $$ F_{AB} = (2 + 1) 	imes 2 = 3 	imes 2 = 6 	ext{ N} $$
   Alternatively, analyzing block A: The net force on A is $F_{applied} - F_{BA} = m_A a$. Since $F_{BA} = F_{AB}$ (Newton's Third Law [NCERT Class 11, Physics Part I, Sec 4.6]), $14 - F_{AB} = 4(2) \Rightarrow F_{AB} = 6 	ext{ N}$.

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