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A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is________ N. (Round off to the Nearest integer) [Take $g = 10 m s^{- 2}]$

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30 N

20 N

40 N

10 N

answer is A.

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Detailed Solution

From FBD shown,

$T = M g - N, R = m g + N = (M + m) g - T$

For no movement of block,

$T \leq μ R \Rightarrow T \leq μ [(M + m) g - T]$

$\Rightarrow T \leq \frac{μ (M + m) g}{1 + μ}$

$\Rightarrow T_{m a x} = \frac{(0.5) (5 + 4) \times 10}{1 + 0.5} = \frac{45}{1.5}$

$∴ T_{\max} = 30 N$

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