Block A of mass m rests on the plank B of mass 3m which is free to slide on a frictionless horizontal surface. The coefficient of friction between the block and plank is 0.2. If a horizontal force of magnitude 2 mg is applied to the plank B, the acceleration of A relative to the plank and relative to the ground respectively, are:

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$0, \frac{g}{2}$

$0, \frac{2 g}{3}$

$\frac{3 g}{5}, \frac{g}{5}$

$\frac{2 g}{5}, \frac{g}{5}$

answer is D.

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Detailed Solution

$\begin{array}{l} f = μmg; ma = μmg \\ a = 0.2 \times g = g / 5 \\ a_{b} = \frac{2 mg - mg / 5}{3 m} = \frac{9 mg}{5 \times 3 m} = \frac{3 g}{5} \end{array}$
Acceleration of A relative to B is $\frac{g}{5} - \frac{3 g}{5} = \frac{- 2 g}{5}$

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