A `10kg` block rests on a `5kg` bracket as shown in figureThe `5kg` breaket rests on a horizontal surface The coefficient of frictionbetween the `10kg` block and the braket on wich it rests are `mu_(s) = 0.40 and mu_(k) = 0.30`

If the `10 kg` block is not to slide on the braket , the corresponding acceleration of the `5 kg` braacket is

Options

A. `1.6ms^(-2)`
B. `0.8ms^(-2)`
C. `1.2ms^(-2)`
D. `2.4ms^(-2)`

Answer

a

Explanation

Assuming system move togather there is no sliding acceleration of the system

`a = (F)/((3+10))= (F)/(15)`
FBD of m
FBD of m`f - F= ma = 10((F)/(15))`
`rArr f = F(1 +(10)/(15)) = (5)/(3) F`
If there is no sliding `F le mu_(s)N`
`F[(5)/(5)] le 0.4 xx 10 xx 10 rArr F le 24 N`
From (i) `a = (F)/(15)= (24)/(15) = 1.6ms^(-2)`