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Two blocks of masses 4 kg (top) and 6 kg (bottom) are in contact on a frictionless surface. A horizontal 30 N force is applied to the 6 kg block. What is the contact force between the blocks?

RankGuru · Accepted Answer

1. **System Acceleration ($a$)**: Treat the two blocks as a single system of mass $M = 4 + 6 = 10$ kg.
   - $a = \frac{F_{ext}}{M} = \frac{30}{10} = 3 	ext{ m/s}^2$.
2. **Isolate one block**: To find the contact force ($N$), look at the $4$ kg block (the one not directly pushed by 30N).
3. **FBD of 4 kg Block**: The only horizontal force on it is the contact force from the 6 kg block.
   - $F_{net} = m_2 a$
   - $N = 4 	ext{ kg} 	imes 3 	ext{ m/s}^2 = 12$ N.
4. **Verification**: Net force on 6 kg block $= 30 - 12 = 18$ N. Acceleration check: $18 / 6 = 3 	ext{ m/s}^2$ (Matches).

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