WBJEE · Physics · Laws of Motion
Three blocks of masses 4 kg. 2 kg. 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of \(14 \mathrm{N}\) is applied on the 4 kg block, the contact force between the \(4 \mathrm{kg}\) and the \(2 \mathrm{kg}\) block will be 
- A \(2 \mathrm{N}\)
- B \(6 \mathrm{N}\)
- C \(8 \mathrm{N}\)
- D \(14 \mathrm{N}\)
Answer & Solution
Correct Answer
(B) \(6 \mathrm{N}\)
Step-by-step Solution
Detailed explanation
We know that, \(F=m a\) \[ a=\frac{F}{m}=\frac{14}{7}=2 \mathrm{ms}^{-2} \] Hence, from the figure \[ 14-N=4 a \] \[ \begin{array}{r} 14-N=8 \\ N=6 \mathrm{N} \end{array} \]
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