A small block of mass 1kg is placed over a plank of mass 5kg. The length of the plank is 8m. Coefficient of friction between the block and the plank is 0.5 and the ground over which plank is placed is smooth. A constants force F = 50N is applied on the plank in horizontal direction. The time after which the block will separate from the plank is

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Detailed Solution

$\begin{array}{l} f_{k} = μ_{k} N \\ f_{k} = 0 .5 (10) = 5 N \end{array}$

$For 1 kg, \sum F = ma$
$\Rightarrow 5 = a_{1} \Rightarrow a_{1} = 5 m / s^{2}$
$\begin{array}{l} For 5 kg, \sum F = ma \\ \Rightarrow 50 - 5 = 5 a \\ \Rightarrow a_{2} = 9 m / s^{2} \\ a_{rel} = a_{2} - a_{1} \\ = 9 - 5 = 4 m / s^{2} \\ S_{rel} = u_{rel} t + \frac{1}{2} a_{rel} t^{2} \end{array}$
$\begin{array}{l} \Rightarrow + 8 = 0 + \frac{1}{2} (4) t^{2} \\ \Rightarrow t = 2 s \end{array}$

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