A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is gently dropped on it. The coefficient of friction between them is mu = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g= 10 ms^(-2), is

A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is ge

1.	A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is gently dropped on it. The coefficient of friction between them is mu = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g= 10 ms^(-2), is
Answer» `0.4 m` 1.2 m 0.6m zero Solution :Forceof Friction` f^(-)mu MG ` ` thereforea = (f)/( m )= (mu mg)/( m)= mu g= 0.5xx 10= 5 MS^(-2)` USING`v^2- u^2= 2aS ` ` 0^2 -2^2 =2 (-5) xx SimpliesS= 0.4m`