A block lying on a horizontal conv− eyor belt moving at a constant velocity receives a velocity 5m/s at t=0 sec. relative to the ground in the direction opposite to the dir− ction of motion of the conveyor. Aftert=4sec,the velocity of the block becomes equal to the velocity of the belt . the coefficient of friction between the block and the belt is 0.2 . then the velocity of the conveyor belt is: (g=10m/s^2 ) (A) 13 m/s (B) −13m/s (C)3m/s (D) 6m/s


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Question Number 40396 by LXZ last updated on 21/Jul/18

$A b l o c k l y i n g o n a h o r i z o n t a l c o n v -$ $e y o r b e l t m o v i n g a t a c o n s t a n t$ $v e l o c i t y r e c e i v e s a v e l o c i t y 5 m / s$ $a t t = 0 s e c . r e l a t i v e t o t h e g r o u n d$ $i n t h e d i r e c t i o n o p p o s i t e t o t h e d i r -$ $c t i o n o f m o t i o n o f t h e c o n v e y o r .$ $A f t e r t = 4 s e c, t h e v e l o c i t y o f t h e$ $b l o c k b e c o m e s e q u a l t o t h e v e l o c i t y$ $o f t h e b e l t . t h e c o e f f i c i e n t o f$ $f r i c t i o n b e t w e e n t h e b l o c k a n d t h e$ $b e l t i s 0.2 . t h e n t h e v e l o c i t y o f t h e$ $c o n v e y o r b e l t i s : (g = 10 m / s^{2})$ $(A) 13 m / s (B) - 13 m / s$ $(C) 3 m / s (D) 6 m / s$
Commented by prakash jain last updated on 22/Jul/18

$Friction force = 0.2 \times m \times g$ $= 2 m$ $acceleration (opposite to relative$ $motion) = - 2 {ms}^{- 2}$ $v v e l o c i t y i g b e l t w r t t o g r o u n d$ $5 + a t = - v$ $5 - 2 \times 4 = - v \Rightarrow v = 3$
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