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A particle of mass 0.2 kg has an intial ...

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)` ) .

Text Solution

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`AB = (h)/(sin 30^(@)) = (0.5)/(1//2)`
AB = 1m
Let f be the force of friction
Net downward force along the plane on the particle moving up the plane is
`F = m g isn 30^(@) + f = m g sin^(@) + mu m g cos 30^(@)`

`:.` Retaradation of particle `(-a) = (F)/(M) = g sin 30^(@) + mu g cos 30^(@)` `= 10 xx (1)/(2) + (1)/(sqrt(3)) xx 10 xx (sqrt(3))/(2)`
`= 10 m//s^(2)`
Let v be the velocity of particle at top
Then `V^(2) = u^(2) + 2` as `V^(2) = (5)^(2) + 2 (-10) xx 1` `V^(2) = 5` `V = sqrt(5)`
V = 2.24m/s
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