A horizontal conveyor belt move with a constant velocity v . A small block is projected with a velocity of 6m/s on it…

Q: A horizontal conveyor belt move 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 . μ =0.3 , g = 10 m/s2. Find v

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Sol: $\large v_{b,c} = v_b + v_c$

$\large v_{b,c} = 6 + v $

f = ? m g = 0.3 × m × 10 = 3m

Retardation a = F/m = 3m/m = 3 m/s2

ur = 6 + v , vr = 0, t = 4 sec , ar = -3m/s2

vr = ur + ar t

0 = (6 + v) – 3 × 4

v = 6 m/s.