A conveyor belt is moving at a constant speed of 2m/s.A box is gently dropped on it.The coefficient of fric9between
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Frictional force oon the box f = µmg
Acceleration in the box a = µg = 5 ms⁻²
v² = u² + 2as
⇒ 0 = 2² + 2 x (5)s
⇒ s = – 2/5 w.r.t. belt
⇒ distance = 0.4 m
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