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A horizontal conveyor belt moves with a ...

A horizontal conveyor belt moves with a constant velocity `V.A` small block is projected with a velocity of `6m//s` on it in a direction opposite to the direction of motion of the belt The block comes to rest relative to the belt in a time `4s. mu =0.3, g =10 .m//s^(2)` Find `V` .

Text Solution

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`|vecV_(b.c)| = V_(b) + V_(c) = 6 + V`
`f = mu mg = 0.3 xx m xx 10 = 3m `
Retardation `a = (F)/(m) = (3m)/(m) = 3 m//s^(2)`
`u_(r) = 6 + V , V =0` , t = 4 sec , `a_(r) = -3 ms^(-2) , V_(r) = u_(r) + a_(r)t `
` 0 = ( 6 + V) - 3 xx 4 , V= 6 m//s `
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