a)$ 65\; N ,\; a_0=1.96 ms^{-2}$ b)$ 40\; N ,\; a_0=2.62 ms^{-2}$ c)$ 40\; N ,\; a_0=1.96 ms^{-2}$ d)$ 65\; N ,\; a_0=2.62 ms^{-2}$

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As the man is stationary on the belt acceleration of man=acceleration of belt

Net force on the man $=ma$

$\qquad=65 \times 1=65 N$

Coefficient of friction $\mu=0.2$

Limiting friction $=\mu mg$

For the man to be stationary with respect to belt accelerating with $'a_0'$ of belt

$ma_0=\mu mg$

$a_0=\mu g$

$\qquad=0.2 \times 9.8 $

$\qquad=1.96 ms^{-2}$

Hence a is the correct answer.

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