Two blocks `A` and `B` are separated by some distance and tied by a string as shown in the figure . The force of friction in both the blocks at `t = 2 s` is

Options

A. `4N(rarr),5N(larr)`
B. `2N(rarr),5N(larr)`
C. `0N(rarr),10N(larr)`
D. `1N(larr),10N(larr)`

Answer

D

Explanation

At `t =2s, F_(1) =4N`
`f_(1) mu_(1) =0.6 f_(2) mu_(2) =0.5`
`f_(1) =mu_(1) m_(1)g =0.6 xx 1 xx 10 =6N` ltbr gt `f_(2) =mu_(2) m_(2)g =0.5 xx 2xx 10 =10N`
`F_("net") =F_(2) -F_(1) =15 -4 =11N`
As `F_("net")gt f_(1) +f_(2)`
The system will remain at rest and the valuse of frictional forces on the blocks will be
`T =4+ f_(1)` and `T =15 -f_(2), 4+ f_(1) =15 -f_(2)`
`rf_(1) +f_(2) =11N rarr(1)`
4th option
`f_(1) =+1N, f_(2) = +10N , f_(1) +f_(2) =11N`
.