A conveyor belt is moving at a constant speed | vedclass

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(A) The box is dropped on the moving belt. Initially,the box is at rest relative to the ground,but the belt is moving at $v = 2\, m s^{-1}$.Relative t

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) The box is dropped on the moving belt. Initially,the box is at rest relative to the ground,but the belt is moving at $v = 2\, m s^{-1}$.Relative to the belt,the box has an initial velocity $u_{rel} = 2\, m s^{-1}$.The kinetic friction force acting on the box is $f = \mu N = \mu mg$.The acceleration of the box relative to the belt is $a = \frac{f}{m} = \mu g = 0.5 	imes 10 = 5\, m s^{-2}$.Since the friction force opposes the relative motion,the box will decelerate relative to the belt until its velocity relative to the belt becomes zero.Using the equation of motion $v_{rel}^2 = u_{rel}^2 - 2as$,where $v_{rel} = 0$:$0^2 = 2^2 - 2(5)s$$10s = 4$$s = 0.4\, m$.

$A$ conveyor belt is moving at a constant speed of $2\, m s^{-1}$. $A$ 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 $g = 10\, m s^{-2}$,is ........... $m$.

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