A force F = Kt (where t is time in seconds and K = 2, N/s ) is applied on 2 ,kg block at t = 0

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4-2.Friction

hard

A force $F = Kt$ (where $t$ is the time in seconds and $K = 2\, N/s$) is applied on $2 \,kg$ block at $t = 0$ as shown in the figure. The displacement of $8\ kg$ block till the time when $2\, kg$ block start slipping on $8\,kg$ block will be (coefficient of friction between $2\,kg$ block and $8\, kg$ block is $0.2$ and between $8\, kg$ block and surface is zero,
$g = 10m/s^2)$

A

$\frac{125}{240}\,m$

B

$\frac{250}{240}\,m$

C

$\frac{125}{120}\,m$

D

$1\,m$

Solution

Let common acceleration be $a \, \mathrm{m} / \mathrm{s}^{2}$

$2 t-F_{S}=2 a$ $…(i)$

$F_{S}=8 a$ $…(ii)$

$F_{S}=\frac{2 t}{10} \times 8 \leq 0.2 \times 2 \times 10$

$t=\frac{5}{2} \mathrm{s}, a=\frac{t}{5}$

$\frac{d v}{d t}=\frac{t}{5} \Rightarrow v=\frac{t^{2}}{10}=\frac{d x}{d t} \Rightarrow x=\frac{t^{3}}{30} \mathrm{m}$

Standard 11

Physics

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(AIPMT-2001)