In the arrangement shown,the 2, kg block is h | vedclass

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(B) Initially,the system is at rest. The tension $T_1$ in the string is equal to the weight of the $3\, kg$ block:$T_1 = m_2 g = 3 	imes 10 = 30\, N$

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Decrease of $12\, N$

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(B) Initially,the system is at rest. The tension $T_1$ in the string is equal to the weight of the $3\, kg$ block:$T_1 = m_2 g = 3 	imes 10 = 30\, N$.When the $2\, kg$ block is released,the system accelerates. Let $a$ be the acceleration of the system. The equation of motion for the $3\, kg$ block is $3g - T_2 = 3a$,and for the $2\, kg$ block is $T_2 - 2g \sin 30^{\circ} = 2a$.Adding these equations: $3g - 2g \sin 30^{\circ} = (3 + 2)a$.$a = \frac{3(10) - 2(10)(0.5)}{5} = \frac{30 - 10}{5} = \frac{20}{5} = 4\, m/s^2$.Now,calculate the new tension $T_2$ using the $3\, kg$ block equation:$T_2 = 3g - 3a = 3(10 - 4) = 3(6) = 18\, N$.The change in tension is $\Delta T = T_2 - T_1 = 18 - 30 = -12\, N$.Therefore,the tension decreases by $12\, N$.

In the arrangement shown,the $2\, kg$ block is held to keep the system at rest. The string and pulley are ideal. When the $2\, kg$ block is set free,by what amount does the tension in the string change? $[g = 10\, m/s^2]$

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