A body of mass 10 kg lies on a rough horizontal surface. When a horizontal force of F newtons acts on it, it gets an acceleration of `5 m//s^(2)`. And when the horizontal force is doubled, it gets an acceleration of `18 m//s^(2)`. The coefficient of friction between the body and the horizontal surface is

Explanation

On a rough horizontal plane, acceleration of a block of mass .m. is given by `a=(P)/(m)-mu_(k) g " "` ….(i)
Initially, `P =F , a=5 ms^(-2)`
`5=(F)/(10) mu_(k)(10) …..(ii) (because m = 10 kg)`
When force is doubled i.e., `P=2F , a = 18 ms^(-2)`.
`18=(2F)/(10)-mu_(K)(10)` .....(iii)
Multiplying Eq. (ii) with 2 and subtracting from Eq. (iii)
`8=mu_(k)(10)rArr mu_(k)=(8)/(10)=0.8`