Block of mass m rests on the plank B of mass 3m which is free to slide on a frictionless horizontal surface. The coefficient of friction between the block and plank is 0.2. If a horizontal force of magnitude 2 mg is applied to the plank B, the acceleration of A relative to the plank and relative to the ground respectively, are

Options

A. `0,(g),(2)`
B. `0,(2g)/(3)`
C. `(3g)/(5),(g)/(5)`
D. `(2g)/(5),(g)/(5)`

Answer

D

Explanation

(d) Block A moves due to friction. Maximum value of friction can be `mum_(A)g`. Therefore, maximum acceleration of A can be `(mu mg)/(m_(a))` or `mug=0.2g=(g)/(5)`. When force 2 mg is applied on lower block, common acceleration (if both move together) will be,

`a=("Net force")/("Total mass")=(2 mg)/(4m)=(g)/(2)`
Since a =g/2 is greater than maximum acceleration of A which can be give to it by friction. Therefore slipping will take place.
`a_(A)=0.2g=(g)/(5)`
`a_(B)=(2mg-0.2 mg)/(3m)=0.6 g=(3g)/(5)`
`a_(AB)=a_(A)-a_(B)=+((g)/(5)-(3g)/(5))`
`=(-2g)/(5)`(in backward direction)