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(D) Let the masses be $m_1 = 8\,kg$ and $m_2 = 2\,kg$. Both blocks are on the same inclined plane with angle $	heta = 30^o$.The acceleration of any b

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(D) Let the masses be $m_1 = 8\,kg$ and $m_2 = 2\,kg$. Both blocks are on the same inclined plane with angle $	heta = 30^o$.The acceleration of any block sliding down a frictionless inclined plane is given by $a = g \sin 	heta$.For both blocks,the acceleration is $a = g \sin 30^o = 10 	imes 0.5 = 5\,m/s^2$.Since both blocks have the same acceleration and are moving together,the string connecting them remains slack or does not experience any pulling force.Alternatively,using the equation of motion for the system: $(m_1 + m_2)g \sin 	heta = (m_1 + m_2)a$,which gives $a = g \sin 	heta$.For the $2\,kg$ block: $m_2 g \sin 	heta - T = m_2 a$.Substituting $a = g \sin 	heta$,we get $m_2 g \sin 	heta - T = m_2 (g \sin 	heta)$,which implies $T = 0$.Therefore,the tension in the string is $0\,N$.

Two blocks of mass $8\,kg$ and $2\,kg$ are connected by a string and they are released on an inclined plane of inclination $30^o$ as shown in the figure. What will be the tension in the string connecting the two blocks (in $,N$)? (Assume the surface is frictionless and $g = 10\,m/s^2$)

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