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