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 ]](Q24168.png)
$$\mathrm{A}\:\mathrm{plank}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{sliding}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{1}} .\:\mathrm{A}\:\mathrm{another}\:\mathrm{block}\:\mathrm{of} \\ $$$$\mathrm{mass}\:{M}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{gently}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{it}.\:\mathrm{The} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plank} \\ $$$$\mathrm{is}\:\mathrm{0}.\mathrm{2}.\:\mathrm{Assuming}\:\mathrm{that}\:\mathrm{plank}\:\mathrm{is}\:\mathrm{long} \\ $$$$\mathrm{enough}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{block}\:\mathrm{does}\:\mathrm{not}\:\mathrm{fall} \\ $$$$\mathrm{from}\:\mathrm{it}.\:\mathrm{The}\:\mathrm{velocity}-\mathrm{time}\:\mathrm{graph}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{block}\:\mathrm{is}\:\left[\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$
Commented by ajfour last updated on 14/Nov/17
Commented by Tinkutara last updated on 14/Nov/17
$$\mathrm{Why}\:\mathrm{it}\:\mathrm{only}\:\mathrm{increases}\:\mathrm{upto}\:\mathrm{2}.\mathrm{5}\:\mathrm{s}? \\ $$
Commented by ajfour last updated on 14/Nov/17
$${from}\:{conservation}\:{of}\:{momentum} \\ $$$${for}\:\left({plank}+{block}\right):\:{we}\:{have} \\ $$$$\:\:{Mu}\:=\mathrm{2}{Mv}_{{final}\:{common}\:{velocity}} \\ $$$$\Rightarrow\:{v}_{{final}\:{common}\:{velocity}} \:=\:\frac{{u}}{\mathrm{2}}=\:\mathrm{5}{m}/{s} \\ $$$${time}\:{taken}\:{for}\:{block}\:{to}\:{reach}\:{this} \\ $$$${velocity}\:{is}\:{obtainable}\:{from}\:: \\ $$$$\:{v}=\mathrm{0}+\mu{gt} \\ $$$$\mathrm{5}\:=\:\left(\mathrm{0}.\mathrm{2}×\mathrm{10}\right){t} \\ $$$$\:{t}=\:\mathrm{2}.\mathrm{5}\:{s}\:. \\ $$$${After}\:{this}\:{they}\:{move}\:{together} \\ $$$${with}\:{the}\:{same}\:{velocity}\:{and}\:{the} \\ $$$${mutual}\:{friction}\:{no}\:{more}\:{acts}. \\ $$
Commented by Tinkutara last updated on 14/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$