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`

The maximum force F that can be applied if the `10 kg` block is not to slide on the braket is

Options

A. `32 N`
B. `24N`
C. `18N`
D. `48N`

Answer

b

Explanation

Assuming system move together 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)`