A system of two blocks is placed on a rought horizontal surface as shown in figure below The coefficient of staic and kinetic friction at two surfaces are shown A force F is horizontal applied on the upper block as shown .Let `f_(1) ,f_(2)` represent the frictional force between upper and lower surface of contact respectively and `a_(1),a_(2)` represent the acceleration of `3kg and 2kg` block respectively

For `F= 12N` mark the correct option

Options

A. `f_(1)= 7.8N,f_(2)= 7.8N, a_(1)= 1.4 ms^(-2), a_(2)= 0ms^(-2)`
B. `f_(1)= 7.8N,f_(2)= 10N, a_(1)= a_(2)= 1.4 ms^(-2), a_(2)`
C. `f_(1)= 7.8N,f_(2)= 5N, a_(1)=a_(2)= 1.4 ms^(-2)`
D. `f_(1)= 7.4N,f_(2)= 5N, a_(1)=a_(2) = 1.2 ms^(-2)`

Answer

c

Explanation

From the data given we can find the limiting friction force for the two surface

`f_(L_1) = 0.5 xx 3 xx 10 = 15 N`
`f_(L_2)= 0.2 xx 5 xx 10 = 10N`
`F lt f_(L_3)`
Both the blocks remain at rest and `f_(1)= F,f_(1) = f_(1)` and `a_(1)` and `a_(2)` For `F gtf_(t_2)` and F is less than a correct value any`F_(1)` the motion stars at lower surface but both the blocks surface continue to move with starts same acceleration The friction on surface becomes kinetic in nature
Here `a = a_(1) = a_(2) = (F- f_(t_2))/(5)ms^(-2)`
`F - f_(1) = 3a` and `f_(1) - f_(k) = 2a`
All these equation give
`f_(1) = (2F- 3f_(t_2))/(5)`
for relative motion to start between two blocks `f_(1) gt f_(t_2) F gt 30 N`so minimum value to F is relative motion between blocks is `30 N` for `f = 12 N,f_(1) = 7.8 N`