A particle is at rest on a rough horizontal plane . The plane is slowly titled until the particle starts to move and is then kept fixed . If the static and kinetic coefficients of friction are `mu_(s)` and `mu_(k) ( lt mu_(s))` respectively , show that the velocity of the particle after it has travelled distance d is given by :
`v = sqrt((2(mu_(s) - mu_(k))"gd")/((1 + mu_(s)^(2))^(1//2)))`

Explanation

Let the inclined plane makes an angle `alpha` with the horizontal . In equilibrium position :
mg cos `alpha - R implies R = "mg cos" alpha`
mg sin `alpha - F_(s) = 0 implies F_(s) = "mg sin" alpha`

`therefore " " mu_(s) = (F_(s))/(R) = "tan" alpha`
During the subsequent motion , the equations of motion are
mg cos `alpha - R = 0 implies R = "mg cos " alpha`
mg sin `alpha - F_(k) = "ma" implies F_(k) = "mg sin " alpha - "ma" `

`therefore " " mu_(k) = (F_(k))/(R) = ( "mg sin" alpha - "ma")/("mg cos" alpha)`
= tan `alpha - (a)/("g cos" alpha)`
= `mu_(s) - (a)/("g cos" alpha)`
or `" " a = (mu_(s) - mu_(k)) "g cos" alpha`
`implies a = ((mu_(s) - mu_(k))g)/((1 + mu_(s)^(2))^(1//2))`
As acceleration of the body sliding down an inclined plane is constant ,
` v^(2) = 2ad = (2(mu_(s) - mu_(k))"gd")/((1 + mu_(s)^(2))^(1//2))`
`therefore " " v = sqrt((2(mu_(s) - mu_(k))"gd")/((1 + mu_(s)^(2))^(1//2)))`