A system of two blocks is shown in figure. Friction coefficient between 5 kg and 10 kg block is `mu=0.6` and between 10 kg and ground is `mu=0.4` What will be the maximum value of force F applied at the lower block so that 5 kg block does not slip w.r.t. 10 kg . `(g=10m//sec^(2)).` The force applied at the upper block is having fixed magnitude of 80 N (both forces start to act simultaneously)

Options

A. 160 N
B. 250 N
C. 210 N
D. 310 N

Explanation

`(F+80N)-mu_(2)(5kg+10kg)g = (15kg)a`
`(F+80-60)/(15)=a` ...(1)

`F-mu_(2)(15)xx10-mu_(1)(5xx10)=10xxa`
`(F-90)/(10)=a` ...(2)
So, we can write
`(F+20)/(15)=(F-90)/(10)`
F=310 N