In the given figure the acceleration of M is | vedclass

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Question

$\sin 	heta=\frac{3}{5} \quad \Rightarrow \cos 	heta=\frac{4}{5}$

$\mathrm{N}=\mathrm{mg}-\mathrm{F} \sin 	heta=\frac{10}{3} 	imes 10-50 	imes

Vedclass · Accepted Answer

$\frac{80}{3} \, ms^{-2}$

Vedclass · Answer

$\sin 	heta=\frac{3}{5} \quad \Rightarrow \cos 	heta=\frac{4}{5}$

$\mathrm{N}=\mathrm{mg}-\mathrm{F} \sin 	heta=\frac{10}{3} 	imes 10-50 	imes \frac{3}{5}=\frac{10}{3}$

and $\mathrm{F}+\mathrm{F} \cos 	heta-\mu \mathrm{N}=\mathrm{Ma}$

$\Rightarrow \quad 50+50 	imes \frac{4}{5}-\frac{1}{3} 	imes \frac{10}{3}=\frac{10}{3} \mathrm{a}$

$\Rightarrow \quad 90-\frac{10}{9}=\frac{10}{3} \mathrm{a} \Rightarrow \mathrm{a}=\frac{80}{3} \mathrm{m} / \mathrm{s}^{2}$

In the given figure the acceleration of $M$ is $(g = 10 \,ms^{-2})$

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