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

Options

A. 0.2
B. 0.8
C. 0.4
D. 0.6

Answer

B

Explanation

`m = 10 kg, a_(1) = 5 ms^(-2), F_(1) =F, a_(2) = 18 ms^(-2), F2= 2F`
From Newton's second law, F -f=m
`1. F-f=ma_(1)`
Cofficient of friction, ` mu = f/R`
` f=muR`
` implies F-muR = ma_(1)`
R = mg
` implies F -mumg = ma_(1)`
` :. F =ma_(1) + mumg`
` 2. F_(2)-f=ma_(2)`
` implies 2F-mumg = ma_(2)`
` :. 2F = mumg+ma_(2)`
substuting equation (1) in (2)
` 2(ma_(1)+mumg)=mumg+ma_(2)`
`implies mumg = ma_(2)-2ma_(1)`
`implies mug = a_(2)-2a_(1)`
`:. mu = (a_(2)-2a_(1))/g = (18 - 2xx5)/10 =0.8`.