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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-




Question Number 40396 by LXZ last updated on 21/Jul/18
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
$${A}\:{block}\:{lying}\:{on}\:{a}\:{horizontal}\:{conv}− \\ $$$${eyor}\:{belt}\:{moving}\:{at}\:\:{a}\:{constant}\: \\ $$$${velocity}\:{receives}\:{a}\:{velocity}\:\mathrm{5}{m}/{s} \\ $$$${at}\:{t}=\mathrm{0}\:{sec}.\:{relative}\:{to}\:{the}\:{ground}\: \\ $$$${in}\:{the}\:{direction}\:{opposite}\:{to}\:{the}\:{dir}− \\ $$$${ction}\:{of}\:{motion}\:{of}\:{the}\:{conveyor}. \\ $$$${Aftert}=\mathrm{4}{sec},{the}\:{velocity}\:{of}\:{the}\: \\ $$$${block}\:{becomes}\:{equal}\:{to}\:{the}\:{velocity} \\ $$$${of}\:{the}\:{belt}\:.\:{the}\:{coefficient}\:{of}\: \\ $$$${friction}\:{between}\:{the}\:{block}\:{and}\:{the} \\ $$$${belt}\:{is}\:\mathrm{0}.\mathrm{2}\:.\:{then}\:{the}\:{velocity}\:{of}\:{the} \\ $$$${conveyor}\:{belt}\:{is}:\:\:\left({g}=\mathrm{10}{m}/{s}^{\mathrm{2}} \right) \\ $$$$\left({A}\right)\:\mathrm{13}\:{m}/{s}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({B}\right)\:\:−\mathrm{13}{m}/{s} \\ $$$$\left({C}\right)\mathrm{3}{m}/{s}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({D}\right)\:\:\:\mathrm{6}{m}/{s} \\ $$
Commented by prakash jain last updated on 22/Jul/18
Friction force=0.2×m×g   =2m  acceleration (opposite to relative  motion)=−2 ms^(−2)   v velocity ig belt wrt to ground  5+at=−v  5−2×4=−v⇒v=3
$$\mathrm{Friction}\:\mathrm{force}=\mathrm{0}.\mathrm{2}×{m}×{g}\: \\ $$$$=\mathrm{2}{m} \\ $$$$\mathrm{acceleration}\:\left(\mathrm{opposite}\:\mathrm{to}\:\mathrm{relative}\right. \\ $$$$\left.\mathrm{motion}\right)=−\mathrm{2}\:\mathrm{ms}^{−\mathrm{2}} \\ $$$${v}\:{velocity}\:{ig}\:{belt}\:{wrt}\:{to}\:{ground} \\ $$$$\mathrm{5}+{at}=−{v} \\ $$$$\mathrm{5}−\mathrm{2}×\mathrm{4}=−{v}\Rightarrow{v}=\mathrm{3} \\ $$

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