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A-plank-of-mass-M-kg-is-sliding-on-the-smooth-horizontal-surface-with-constant-velocity-of-10-ms-1-A-another-block-of-mass-M-kg-is-gently-placed-on-it-The-coefficient-of-friction-between-the-blo




Question Number 24168 by Tinkutara last updated on 13/Nov/17
A plank of mass M kg is sliding on the  smooth horizontal surface with constant  velocity of 10 ms^(−1) . A another block of  mass M kg is gently placed on it. The  coefficient of friction between the  block and the upper surface of the plank  is 0.2. Assuming that plank is long  enough such that the block does not fall  from it. The velocity-time graph of the  block is [Take g = 10 m/s^2 ]
$$\mathrm{A}\:\mathrm{plank}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{sliding}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{1}} .\:\mathrm{A}\:\mathrm{another}\:\mathrm{block}\:\mathrm{of} \\ $$$$\mathrm{mass}\:{M}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{gently}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{it}.\:\mathrm{The} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plank} \\ $$$$\mathrm{is}\:\mathrm{0}.\mathrm{2}.\:\mathrm{Assuming}\:\mathrm{that}\:\mathrm{plank}\:\mathrm{is}\:\mathrm{long} \\ $$$$\mathrm{enough}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{block}\:\mathrm{does}\:\mathrm{not}\:\mathrm{fall} \\ $$$$\mathrm{from}\:\mathrm{it}.\:\mathrm{The}\:\mathrm{velocity}-\mathrm{time}\:\mathrm{graph}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{block}\:\mathrm{is}\:\left[\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$
Commented by ajfour last updated on 14/Nov/17
Commented by Tinkutara last updated on 14/Nov/17
Why it only increases upto 2.5 s?
$$\mathrm{Why}\:\mathrm{it}\:\mathrm{only}\:\mathrm{increases}\:\mathrm{upto}\:\mathrm{2}.\mathrm{5}\:\mathrm{s}? \\ $$
Commented by ajfour last updated on 14/Nov/17
from conservation of momentum  for (plank+block): we have    Mu =2Mv_(final common velocity)   ⇒ v_(final common velocity)  = (u/2)= 5m/s  time taken for block to reach this  velocity is obtainable from :   v=0+μgt  5 = (0.2×10)t   t= 2.5 s .  After this they move together  with the same velocity and the  mutual friction no more acts.
$${from}\:{conservation}\:{of}\:{momentum} \\ $$$${for}\:\left({plank}+{block}\right):\:{we}\:{have} \\ $$$$\:\:{Mu}\:=\mathrm{2}{Mv}_{{final}\:{common}\:{velocity}} \\ $$$$\Rightarrow\:{v}_{{final}\:{common}\:{velocity}} \:=\:\frac{{u}}{\mathrm{2}}=\:\mathrm{5}{m}/{s} \\ $$$${time}\:{taken}\:{for}\:{block}\:{to}\:{reach}\:{this} \\ $$$${velocity}\:{is}\:{obtainable}\:{from}\:: \\ $$$$\:{v}=\mathrm{0}+\mu{gt} \\ $$$$\mathrm{5}\:=\:\left(\mathrm{0}.\mathrm{2}×\mathrm{10}\right){t} \\ $$$$\:{t}=\:\mathrm{2}.\mathrm{5}\:{s}\:. \\ $$$${After}\:{this}\:{they}\:{move}\:{together} \\ $$$${with}\:{the}\:{same}\:{velocity}\:{and}\:{the} \\ $$$${mutual}\:{friction}\:{no}\:{more}\:{acts}. \\ $$
Commented by Tinkutara last updated on 14/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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