A block is gently placed at the top of an inclined plane 6.4 m long . Find the time taken by the block to slide down to the bottom of the plane . The plane makes an angle` 30^(@)` with the horizontal Coefficient of friction between the block and the plane is 0.2 Take ` g = 10 m//s^(2) ` .

Explanation

As the block is sliding down the inclined plane force of friction F acts up the plane If a is acceleration produced in the block then net force on the block down the plane is
` f = ma = mg sin theta - F = mg sin theta - mu R `
` = mg sin theta - mu mg cos theta `
`= mg (sin theta - mu cos theta)`
or `a = g (sin theta- mu cos theta ) `
` = 10 (sin 30^(@) - 0.2 cos 30^(@)) `
` = 10 ((1)/(2) - 0 .2 sqrt(3)/(2)) `
` a = 5 (1 - 0.2 xx 1.732) = 5 xx 0 .6536 `
` = 3 . 268 m//s^(2) `
Now ` s= 6.4 m`,
` u = 0 ` as the block starts from rest
` t = ? `
From ` s = ut + (1)/(2) at^(2) `
`6.4 = 0 + (1)/(2) xx 3 . 268 t^(2)`
` t = sqrt((2 xx 6.4 )/(3.268))= 1.98 s `
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