Two blocks `m_(1) = 4 kg` and `m_(2) = 2` kg connected by a weightless rod slide down a plane having an inclination of `37^(@)` . The coefficient of dynamic friction of `m_(1)` and `m_(2)` with the inclined plane are `mu_(1) = 0 .75` and `mu_(2) = 0 . 25` respectively Find the common acceleration of the two blocks and tension in the rod Take is `37^(@) = 0.6` and `cos 37^(@) = 0 .8` .

Explanation

Let a be the common acceleration of the blocks and T be the tension in the connecting Components of wts of two blocks down the inclined plane
`= m_(1) g sin 37^(@) + m_(2) g sin 37^(@) `
` = (m_(1) + m_(2)) g sin 37^(@) `
Frictional forces on two blocks up the inclined
plane ` = mu_(1) m_(1) g cos 37^(@) + mu_(2) m_(2) g cos 37^(@) `
` (mu_(1) m_(1) + mu_(2) m_
(2)) g cos 37^(@) `
Net force down the plane ` = (m_(1) + m_(2)) a `
` :. (m_(1) + m_(2)) a = (m_(1) + m_(2) ) g sin 37^(@)`
` -(mu_(1) m_(1) + mu_(2) m_(2)) g cos 37^(@)`
Putting the valuse , we get ` a = 1.3 m//s^(2) `
To find T , we write eqn . of motion of block `m_(1)`
` T + m_(1) g sin 37^(@) - mu_(1) m_(1) g cos 37^(@)= ma `
on solving we get `T = 5.2 N ` .