Two blocks of masses 1 kg and 2 kg are conn | vedclass

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(B) The frictional force acting on both blocks is kinetic in nature.For the $1 \ kg$ block:$f_1 = \mu m_1 g \cos 45^{\circ} = 0.4 	imes 1 	imes 10 \

Vedclass · Accepted Answer

$3$

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(B) The frictional force acting on both blocks is kinetic in nature.For the $1 \ kg$ block:$f_1 = \mu m_1 g \cos 45^{\circ} = 0.4 	imes 1 	imes 10 	imes \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ N$For the $2 \ kg$ block:$f_2 = \mu m_2 g \cos 45^{\circ} = 0.4 	imes 2 	imes 10 	imes \frac{1}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \ N$The net force along the incline in the downward direction provides the acceleration $a = \alpha \sqrt{2}$:$(m_1 + m_2)g \sin 45^{\circ} - (f_1 + f_2) = (m_1 + m_2)a$$(1 + 2) 	imes 10 	imes \frac{1}{\sqrt{2}} - (2\sqrt{2} + 4\sqrt{2}) = (1 + 2) 	imes (\alpha \sqrt{2})$$30 	imes \frac{1}{\sqrt{2}} - 6\sqrt{2} = 3\alpha \sqrt{2}$Multiply throughout by $\sqrt{2}$:$30 - 6 	imes 2 = 3\alpha 	imes 2$$30 - 12 = 6\alpha$$18 = 6\alpha$$\alpha = 3$

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