A particle of mass 0.2 kg has an intial speed of `5 ms^(-1) ` at the bottom of a rough inclined plane of inclination `30^(@)` and vertical height 0.5 m . What is the speed of the particle as it reaches the top of the inclined plan ? (Take `mu = 1//sqrt3 , g = 10 ms^(-2)` ) .

Explanation

Here, `m = 0.2 kg, u = 5 m//s, 0 = 30^(@), h = 0.5 m`
Here, length `AB = (h)/(sin 30^(@)) = (0.5)/(1//2) = 1m`.
Lef f be the force of faction
Net downward force along the plane on the particle moving up the plane is `F = mg sin 30^(@) + f = mg sin 30^(@) + mu mg cos 30^(@)`
` :.` Retardation of particle `(-a) = (F)/(m) = g sin 30^(@) + mu g cos 30^(@) = 10 xx (1)/(2) + (1)/sqrt3 xx 10 xx sqrt3/(2) = 10 m//s^(2)`
Let upsilon be the velocity of particle at B. Using the relation `upsilon^(2) = u^(2) + 2`as
We have `upsilon^(2) =5^(2) +2 (-10) xx 1=5` or `upsilon= sqrt5 m//s = 2.24 ms^(-1)`
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