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A plank of mass 2kg and length 1m is placed on horizontal floor. A small block of mass 1kg is placed on top of the plank, at its right extreme end. The coefficient of friction between plank and floor is 0.5 and that between plank and block is 0.2. If a horizontal force =30N starts acting on the plank to the right, the time after which the block will fall off the plank is (Take, $g = 10 {ms}^{- 2}$ )

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$(\frac{2}{3}) s$

$15 s$

$0.75 s$

$(\frac{4}{3}) s$

answer is A.

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Detailed Solution

${(f_{1})}_{\max} = between 1 kg and 2 kg$

$= 0.2 \times 1 \times 10 = 2 N {(f_{2})}_{\max} = between 2 kg and ground$

$\to 2 N 15^{'} N = 0.5 \times 3 \times 10 = 15 N a_{1} = \frac{2}{1} = 2 m / s^{2} a_{2} = \frac{30 - 15 - 2}{2} = 6.5 m / s^{2} a_{r} = a_{2} - a_{1} = 4.5 m / s^{2} t = \sqrt{\frac{2 S_{r}}{a_{r}}} = \sqrt{\frac{2 \times 1}{4.5}} = \frac{2}{3} s$

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