Two blocks which are connected to each other | vedclass

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Question

$a_{1}=a_{2}=\frac{m g \sin 53^{\circ}-\operatorname{mg} \sin 37^{\circ}}{2 m}=\frac{g}{10}=1 \mathrm{m} / \mathrm{s}^{2}$

$a_{c m}=\frac{m \sqrt{a

Vedclass · Accepted Answer

$\frac{1}{\sqrt 2}\,m/s^2$

Vedclass · Answer

$a_{1}=a_{2}=\frac{m g \sin 53^{\circ}-\operatorname{mg} \sin 37^{\circ}}{2 m}=\frac{g}{10}=1 \mathrm{m} / \mathrm{s}^{2}$

$a_{c m}=\frac{m \sqrt{a_{1}^{2}+a_{2}^{2}}}{2 m}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{m} / \mathrm{s}^{2}$

Two blocks which are connected to each other by means of a massless string are placed on two inclined planes as shown in figure. After releasing from rest, the magnitude of acceleration of the centre of mass of both the blocks is $(g = 10\, m/s^2)$

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