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Two blocks which are connected to each other by means of massless string are placed on two inclined planes as shown in figure.After releasing from rest the magnitude of the acceleration of the centre of mass of both the blocks

Two blocks which are connected to each other by means of massless string are placed on two inclined planes as shown in figure.After releasing from rest the magnitude of the acceleration of the centre of mass of both the blocks

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Grade:12th pass

1 Answers

rudransh
18 Points
5 years ago
Since normal reaction will nt provide any force here,
The effective force on the particle on the particle on the right = mg sin\Theta = mg sin 53 =(4/5) mg 
The effective force on the particle on the particle on the left = mg sin\Theta = mg sin 37 =(3/5) mg 
\therefore total force = (force on left) – (force on right)=mg(1/5)
\becauseF=ma
\therefore a=F/m=\frac{(1/5)mg}{2m} =g/10=1m/^{s^{2}}
hence, answer is (1)1m/^{s^{2}}
 

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