If M_1 = M_2 = 5 , kg and theta = 30^circ,the | vedclass

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(A) For a system where mass $M_1$ is on an inclined plane and $M_2$ is hanging vertically,the acceleration $a$ of the system is given by:$a = \frac{M_

Vedclass · Accepted Answer

$37.5$

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(A) For a system where mass $M_1$ is on an inclined plane and $M_2$ is hanging vertically,the acceleration $a$ of the system is given by:$a = \frac{M_2 g - M_1 g \sin 	heta}{M_1 + M_2}$Substituting the given values $M_1 = 5 \, kg$,$M_2 = 5 \, kg$,$	heta = 30^\circ$,and taking $g = 10 \, m/s^2$:$a = \frac{5 	imes 10 - 5 	imes 10 	imes \sin 30^\circ}{5 + 5} = \frac{50 - 25}{10} = 2.5 \, m/s^2$Now,for the hanging mass $M_2$,the equation of motion is $M_2 g - T = M_2 a$,where $T$ is the tension in the string.$T = M_2(g - a) = 5(10 - 2.5) = 5 	imes 7.5 = 37.5 \, N$.

If $M_1 = M_2 = 5 \, kg$ and $\theta = 30^\circ$,then the tension in the string will be ........... $N$.

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