Question

A box is kept on a rough horizontal surface.A horizontal force just strong enough to move the box is applied.This force is maintained for 2 seconds and is then removed.The total distance moved by the box is S meter.Then find the value of 6S. Take mu_(S)=0.2& mu_(K)=0.15​

Answer 1

Given:

A box is kept on a rough horizontal surface.A horizontal force just strong enough to move the box is applied.This force is maintained for 2 seconds and is then removed.The total distance moved by the box is S meter.

To find:

Value of 6S ?

Diagram:

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(15,15){\sf{m}}}\put(-20,-0.1){\line(1,0){55}}\multiput(-19,0)(3,0){19}{\qbezier(0,0)(-1,-1)(-2,-2)}\put(15,7.5){\vector(1,0){15}}\put(7.5,0){\vector(0,-1){15}}\put(4,-20){\sf{mg}}\put(7.5,15){\vector(0,1){15}}\put(6.3,31.5){\sf{N}}\put(0,0.2){\vector(-1,0){15}}\put(-17,1.3){\sf{f}}\put(32,6){\sf F}\end{picture}$

Calculation:

Since the applied force is just adequate to move the box , the force should be equal to the static friction :

$\therefore \: F = \mu_{s}(mg) \: \: \: \: \: \: \: .........(1)$

Now , according to free body digram of block:

$\therefore \: F - \mu_{k}(mg) = ma$

$\implies \: \mu_{s}(mg) - \mu_{k}(mg) = ma$

$\implies \: \mu_{s}(g) - \mu_{k}(g) = a$

$\implies \: a = \mu_{s}(10) - \mu_{k}(10)$

$\implies \: a = 0.2(10) - 0.15(10)$

$\implies \: a = 2 - 1.5$

$\implies \: a = 0.5 \: m {s}^{ - 2}$

Now , let Displacement be S:

$\therefore \: S= ut + \dfrac{1}{2} a {t}^{2}$

$\implies \: S= 0 + \dfrac{1}{2} (0.5) {(2)}^{2}$

$\implies \: S= \dfrac{1}{2} (0.5) 4$

$\implies \: S= 1 \: metre$

$\implies \: 6S= 6 \: metre$

Value of 6S is 6 metres.