For the system shown in the figure,the pulley | vedclass

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(D) Let the acceleration of the system be $a$ and the tension in the string be $T$.For the block of mass $m$ on the inclined plane,the equation of mot

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$\frac{1}{2} mg \sin 	heta$

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(D) Let the acceleration of the system be $a$ and the tension in the string be $T$.For the block of mass $m$ on the inclined plane,the equation of motion is: $mg \sin 	heta - T = ma$ (Equation $1$).For the block of mass $m$ on the horizontal surface,the equation of motion is: $T = ma$ (Equation $2$).Adding Equation $1$ and Equation $2$,we get: $mg \sin 	heta = 2ma$.Thus,the acceleration of the system is $a = \frac{g \sin 	heta}{2}$.Substituting the value of $a$ in Equation $2$,we get the tension $T = m \left( \frac{g \sin 	heta}{2} ight) = \frac{1}{2} mg \sin 	heta$.

For the system shown in the figure,the pulleys are light and frictionless. The tension in the string will be

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