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

A horizontal converyor belt moves with a constant velocity V. A small

1.	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
Answer» SOLUTION :`\|vec(V)_(b,c)\|=V_(b)+V_(c )=6+V` `f=mu mg = 0.3 xx m xx 10 = 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 = 6 m/s.