Given M_1 = 10 kg,M_2 = 5 kg,theta = 30^o,a | vedclass

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(D) For the given system,the block $M_1$ is on an inclined plane and $M_2$ is hanging vertically.The forces acting on the system are:$1$. For block $M

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Zero

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(D) For the given system,the block $M_1$ is on an inclined plane and $M_2$ is hanging vertically.The forces acting on the system are:$1$. For block $M_1$: The component of weight down the incline is $M_1 g \sin 	heta$.$2$. For block $M_2$: The weight acting downwards is $M_2 g$.The net force $F_{net}$ on the system is $M_2 g - M_1 g \sin 	heta$ (assuming $M_2$ moves downwards).The total mass of the system is $M_1 + M_2$.Using Newton's second law,$a = \frac{F_{net}}{M_{total}} = \frac{M_2 g - M_1 g \sin 	heta}{M_1 + M_2}$.Substituting the given values:$a = \frac{5 	imes 10 - 10 	imes 10 	imes \sin 30^o}{10 + 5}$$a = \frac{50 - 100 	imes 0.5}{15}$$a = \frac{50 - 50}{15} = 0 \ m/s^2$.Thus,the acceleration of the system is zero.

Given $M_1 = 10 \ kg$,$M_2 = 5 \ kg$,$\theta = 30^o$,and $g = 10 \ m/s^2$. What is the acceleration of the system?

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