Laws of Motion - Result Question 22
22. Three blocks A, B and C of masses $4 kg, 2 kg$ and $1 kg$ respectively, are in contact on a frictionless surface, as shown. If a force of $14 N$ is applied on the $4 kg$ block then the contact force between A and B is
[2015]
(a) $6 N$
(b) $8 N$
(c) $18 N$
(d) $2 N$
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Answer:
Correct Answer: 22. (a)
Solution:
- (a) Acceleration of system, $a=\frac{F _{\text{net }}}{M _{\text{total }}}$ $=\frac{14}{4+2+1}=\frac{14}{7}=2 m / s^{2}$
The contact force between $A$ and $B$
$ =(m_B+m_C) \times a=(2+1) \times 2=6 N $
