A conveyor belt is moving at a constant speed | vedclass

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(A) The box is dropped on the moving belt,so its initial velocity relative to the belt is $u_{rel} = 2\, ms^{-1}$.The force of friction acting on the

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) The box is dropped on the moving belt,so its initial velocity relative to the belt is $u_{rel} = 2\, ms^{-1}$.The force of friction acting on the box is $f = \mu mg$.According to Newton's second law,the acceleration of the box relative to the belt is $a = \frac{f}{m} = \mu g$.Substituting the given values,$a = 0.5 	imes 10 = 5\, ms^{-2}$.We use the kinematic equation $v^2 - u^2 = 2aS$ to find the distance $S$ covered by the box relative to the belt until it stops sliding.Here,the final relative velocity $v = 0\, ms^{-1}$ and the initial relative velocity $u = 2\, ms^{-1}$.Substituting these into the equation: $0^2 - 2^2 = 2(-5)S$.$-4 = -10S$.$S = \frac{4}{10} = 0.4\, m$.

$A$ conveyor 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 the belt before coming to rest on it,(taking $g = 10\, ms^{-2}$) is ........ $m$.

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