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A horizontal converyor belt moves with a constant velocity V. A small block is projected with a velocity of 6 m/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» SOLUTION :`|vec(V)_(b,c)|=V_(b)+V_(c )=6+V` `f=mu mg=0.3xx m xx10=3M` Retardation `a=(3m)/(m)=3m//s^(2)` `u_(r )=6+V, V_(r )=0, t=4 sec` `a_(r )=-3 MS^(-2)` `V_(r ) =u_(r )+a_(r )t` `0=(6+V)-3xx4` `V=6m//s` |
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