A plank A of mass `M` rests on a smooth horizontal surface over which it can move without friction A cabe B of mass m lies on the plank at one edge .The coefficient of friction between the plank and the cube is `mu` The size of cube is very small in comparison to the plank.

At what force `F` applied to the plank in the horizontal direction will be cube begin to slide towards the other end of the plank?

Options

A. `sqrt((Ml)/(F - mu g(M+m)))`
B. `sqrt((2Ml)/(F - mu g(M+m)))`
C. `sqrt((Ml)/(F + mu g(M+m)))`
D. `sqrt((Ml)/(F + mu g(M+m)))`

Answer

b

Explanation

Free- body diagram

Let both the blocks move together
Acceleration of blocks `a = (F)/((m + M))`
`f = m((F)/(m+M))`
if both the blocks move togather` fle mu mg`
`(mF)/((m+M)) le mu mg`
`F le mu (m +M)g`
if the cube begain to slide `F = mu(m+M)g`
`a_(m) = (f)/(m) = mu g`(toward +x direction)
`a_(M) = (F-mu mg)/(M)` (toward +x direction)
`vec a_(m,M) = vec a_(m) - vec a_(M) = mu g - ((F- mu mg)/(M))`
`= (mu(m + M)g- F)/(M)`
if the cube falls from the plank it will cover a distance`1`
`- 1 = (1)/(2) a_(mM)t^(2) rArr t = sqrt((2l)/(a_(M,m))) = sqrt((2lM)/(F - mu(m+M)g))`