In the given figure,the acceleration of the b | vedclass

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(D) Given: $M = \frac{10}{3} \, kg$,$F = 50 \, N$,$\mu = \frac{1}{3}$,$	heta = \sin^{-1}(\frac{3}{5})$.From $\sin 	heta = \frac{3}{5}$,we get $\cos

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(D) Given: $M = \frac{10}{3} \, kg$,$F = 50 \, N$,$\mu = \frac{1}{3}$,$	heta = \sin^{-1}(\frac{3}{5})$.From $\sin 	heta = \frac{3}{5}$,we get $\cos 	heta = \frac{4}{5}$.The vertical forces on the block are the normal force $N$,the weight $Mg$,and the vertical component of the applied force $F \sin 	heta$ (acting upwards).$N + F \sin 	heta = Mg \Rightarrow N = Mg - F \sin 	heta$.$N = (\frac{10}{3} 	imes 10) - (50 	imes \frac{3}{5}) = \frac{100}{3} - 30 = \frac{100 - 90}{3} = \frac{10}{3} \, N$.The horizontal forces are the horizontal component of the applied force $F \cos 	heta$ and the friction force $f = \mu N$.The equation of motion is $F \cos 	heta - \mu N = Ma$.$50 	imes \frac{4}{5} - \frac{1}{3} 	imes \frac{10}{3} = \frac{10}{3} a$.$40 - \frac{10}{9} = \frac{10}{3} a$.$\frac{360 - 10}{9} = \frac{10}{3} a \Rightarrow \frac{350}{9} = \frac{10}{3} a$.$a = \frac{350}{9} 	imes \frac{3}{10} = \frac{35}{3} \, ms^{-2}$.

In the given figure,the acceleration of the block of mass $M = \frac{10}{3} \, kg$ is (given $g = 10 \, ms^{-2}$,$\mu = \frac{1}{3}$,$F = 50 \, N$,and $\theta = \sin^{-1}(\frac{3}{5})$):

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