Two blocks of masses 5 kg and 3 kg are placed in contact over an inclined surface of angle `37^(@)`, as shown. `mu_(1)` is friction coefficient between 5 kg block and the surface of the incline and similarly, `mu_(2)` is friction coefficient betweem the 3 kg block and the surface of the incline. After the release of the blocks from the inclined surface.

Options

A. a. If `mu_(1) = 0.5` and `mu_(2) = 0.3` then 5 kg block exerts 3 N force on the 3 kg block
B. b. If `mu_(1) = 0.5` and `mu_(2) = 0.3` then `5 kg` block exerts 8 N force on the 3 kg block
C. c. If `mu_(1) = 0.3` and `mu_(2) = 0.5` then 5 kg block exerts 1 N on the 3 kg block
D. d. If `mu_(1) = 0.3` and `mu_(2) = 0.5` then 5 kg block exerts no force on the 3 kg block.

Answer

A, D

Explanation

Case I : `mu_(1) = 0.5, mu_(2) = 0.3`
Along the incline, acceleration of 5 kg block will be less than acceleration of 3 kg block provided they move alone on the incline. The reason is greater friction coefficient of 5 kg block, as acceleration along the incline is `g = sin theta - mug cos theta`
One to the contain, both blocks will move together. In this case FBDs of both are shown.

For 5 kg block
`m_(1)g sin theta+N-mu_(1)m_(1)g cos theta =m_(1) a`
For `3 kg` block
`m_(2)g sin theta - N-mu_(2)m_(2)g cos theta = m_(2)a`.