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Question Number 99529 by Dwaipayan Shikari last updated on 21/Jun/20

Answered by mr W last updated on 21/Jun/20

a=acceleration of m  A=acceleration of M  ma=−mg(μ/2)  ⇒a=−((μg)/2)  MA=−(M+m)gμ+mg(μ/2)  ⇒A=−(1+(m/(2M)))gμ  Δa=a−A=(1/2)(1+(m/M))μg  t=(√((2L)/(Δa)))  ⇒t=2(√(L/((1+(m/M))μg)))

$${a}={acceleration}\:{of}\:{m} \\ $$$${A}={acceleration}\:{of}\:{M} \\ $$$${ma}=−{mg}\frac{\mu}{\mathrm{2}} \\ $$$$\Rightarrow{a}=−\frac{\mu{g}}{\mathrm{2}} \\ $$$${MA}=−\left({M}+{m}\right){g}\mu+{mg}\frac{\mu}{\mathrm{2}} \\ $$$$\Rightarrow{A}=−\left(\mathrm{1}+\frac{{m}}{\mathrm{2}{M}}\right){g}\mu \\ $$$$\Delta{a}={a}−{A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{{m}}{{M}}\right)\mu{g} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{L}}{\Delta{a}}} \\ $$$$\Rightarrow{t}=\mathrm{2}\sqrt{\frac{{L}}{\left(\mathrm{1}+\frac{{m}}{{M}}\right)\mu{g}}} \\ $$

Commented by Dwaipayan Shikari last updated on 21/Jun/20

You are right! sir.

Commented by Dwaipayan Shikari last updated on 21/Jun/20

I want to thank you for spending your valuable time

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