A) 1.25\[m{{s}^{-2}}\]
B) 1.96 \[m{{s}^{-2}}\]
C) 2.5 \[m{{s}^{-2}}\]
D) 3.6 \[m{{s}^{-2}}\]
Correct Answer: B
Solution :
As the man is standing stationary w.r.t. the horizontal conveyor belt, he is also accelerating as 1\[m{{s}^{-2}}\], the acceleration of the belt. Thus, Acceleration of the man, \[a=1\text{ }m{{s}^{-2}}\] Mass of the man, \[M=65\text{ }kg\] Therefore, net force on the man, \[Ma=65\times 1=65\text{ }N\] The limiting friction between the shoes of the man and the belt is given by \[F=\mu Mg=0.2\times 65\times 9.8\text{ }N\] If, the man can remain stationary upto an acceleration say a?, then \[Ma=F\] \[a'\,\,=\,\,\frac{F}{M}\,\,=\,\,\frac{0.2\times 65\times 9.8}{65}\,\,=\,\,1.96\,m{{s}^{-2}}\]
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