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Fig. shows a man standing stationary with respect
to a horizontal conveyer belt that is accelerating with 1 m/s². What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is 0.2, upto what acceleration of the belt can the man continue to be stationary relative to the belt ? (mass of the man is 65 kg).

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$60 N, 1 \cdot 96 m / s^{2}$

$65 N, 1.96 m / s^{2}$

$65 N, 5 \cdot 4 m / s^{2}$

$60 N, 5 \cdot 4 m / s^{2}$

answer is B.

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Detailed Solution

short cut a_max = $μ g$ =0.2 $\times 9.8 = 1.96$

Here acceleration of man is equal to the acceleration of belt, i.e. 1m/s².

Net force on the man,

$F^{'} = ma = 65 \times 1 = 65 N$

The limiting friction between the shoes of the man and the belt is given by

$F = μR = μmg = 0 \cdot 2 \times 65 \times 9 \cdot 8 = 127 \cdot 4 N$

Let the man can remain stationary upto an acceleration a' , then

$\begin{array}{l} F = {ma}^{'} or a^{'} = F / m \\ ∴ a^{'} = \frac{127 \cdot 4}{65} \equiv 1 \cdot 96 m / s^{2} \end{array}$

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