A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 k g is…

Solution

$v=v_0{e}^{-\frac{t}{\tau }}$
$v_0=\frac{J}{m}=2.5 m/s$
$v=v_0{e}^{-\frac{t}{\tau }}$
$\frac{dx}{dt}=v_0{e}^{-\frac{t}{\tau }}$
${\int }_0^{x}dx=v_0{\int }_0^{\tau }{e}^{-\frac{t}{\tau }}dt$
$\int e^{-x}dx=\frac{e^{-x}}{-1}$
$x=v_0{\left[\frac{e^{-\frac{t}{\tau }}}{-\frac{1}{\tau }}\right]}_0^{\tau }$
$x=2.5\left(-4\right)\left(e^{-1}-e^0\right)$
$x=25\left(-4\right)\left(0.37-1\right)$
$x=6.30$

Asked in: JEE Advanced 2018 (Paper 2)

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