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4-2.Friction
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A car is moving along a straight horizontal road with a speed ${v_0}$. If the coefficient of friction between the tyres and the road is $\mu $, the shortest distance in which the car can be stopped is
A$\frac{{v_0^2}}{{2\mu g}}$
B$\frac{{{v_0}}}{{\mu g}}$
C${\left( {\frac{{{v_0}}}{{\mu g}}} \right)^2}$
D$\frac{{{v_0}}}{\mu }$
Solution
(a) Retarding force $F = ma = \mu \,R = \mu \;mg$
$\therefore a = \mu g$
Now from equation of motion ${v^2} = {u^2} – 2as$
$ \Rightarrow \;0 = {u^2} – 2as$
$⇒$ $s = \frac{{{u^2}}}{{2a}} = \frac{{{u^2}}}{{2\mu \;g}}$
$\therefore s= \frac{{v_0^2}}{{2\mu \,g}}$
$\therefore a = \mu g$
Now from equation of motion ${v^2} = {u^2} – 2as$
$ \Rightarrow \;0 = {u^2} – 2as$
$⇒$ $s = \frac{{{u^2}}}{{2a}} = \frac{{{u^2}}}{{2\mu \;g}}$
$\therefore s= \frac{{v_0^2}}{{2\mu \,g}}$
Standard 11
Physics
