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A block of mass 0.5 kg is placed over a long plane of mass 1 kg. The friction coefficient between block and plank is 0.1. The system is placed over a smooth horizontal surface. A time varying force F = 5t newton starts acting on the block as shown in figure. ${(g =10 m/s}^{2})$

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Block will break off the plank before relative motion starts.

Block will break of the plank after the start of relative motion

Relative motion between block and plank starts at t = 15/89 sec

Relative motion between block and plank starts at t = 11/89 sec.

answer is B, C.

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Detailed Solution

$3 t + R = 5$

$F o r b r e a k i n g o f f$

$t = 5 / 3 \sec$

$f_{e} = μ R = \frac{1}{10} [5 - 3 t] = \frac{1}{2} - \frac{3}{10} t .$

$F o r 1.0 k g :$

$f_{e} = (1) a \Rightarrow a = \frac{1}{2} - \frac{3}{10} t$

$F o r 0.5 k g$

$4 t - f_{e} = 0.5 \times a$

$4 t - [\frac{1}{2} - \frac{3}{10} t] = \frac{1}{2} [\frac{1}{2} - \frac{3}{10} t]$

$t = \frac{15}{89} \sec$

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