A particle of mass 0.1kg has an iniital speed of `4ms^(-1)` at a point A on a roudh horizontal road. The coefficient of friction between the object and road is 0.15. The particle moves to a point B at a distance of 2m from A. What is the speed of particle at B ?
Take `g=10ms^(-2)`

Explanation

Here, ` m = 0.1 kg, u = 4ms^(-1), mu = 0.15, s = 2 m, upsilon = ` ?
Retardation due to friction ` (-a) = (mumg)/(m) = mug = 0.15 xx 10 = 1.5 m//s^(2) `
Let upsilon be the velocity of particle at B. Using the relation
` upsilon^(2) =u^(2) + 2 as `
` upsilon^(2) = 4^(2) + 2 (-1.5) xx 2=10 , or upsilon = sqrt10 = 3.16 m//s ` .