A horizontal force $10 ext{ N}$ is applied to a block A as shown in figure. The mass of blocks A and B are $

Question

A horizontal force $10 	ext{ N}$ is applied to a block A as shown in figure. The mass of blocks A and B are $2 	ext{ kg}$ and $3 	ext{ kg}$, respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:

Accepted Answer

1. **Analyze the System:** Since the blocks are in contact and moving on a frictionless surface under a single external force, they move together with a common acceleration ($a$).
2. **Calculate Acceleration:** Consider blocks A and B as a single system.
   - Total mass $M = m_A + m_B = 2 	ext{ kg} + 3 	ext{ kg} = 5 	ext{ kg}$.
   - Net external force $F = 10 	ext{ N}$.
   - According to Newton's Second Law:
     $$ a = \frac{F}{M} = \frac{10}{5} = 2 	ext{ m/s}^2 $$
     [Source 32]
3. **Calculate Contact Force:** Isolate block B. The only horizontal force acting on it is the contact force exerted by block A ($F_{AB}$).
   - Apply Newton's Second Law to block B:
     $$ F_{AB} = m_B 	imes a $$
     $$ F_{AB} = 3 	ext{ kg} 	imes 2 	ext{ m/s}^2 = 6 	ext{ N} $$
   - Alternatively, considering block A: $F - F_{BA} = m_A a \implies 10 - F_{BA} = 2(2) \implies F_{BA} = 6 	ext{ N}$. By Newton's Third Law, the force A exerts on B is equal in magnitude. [Source 110, 115]

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