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4-2.Friction
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The retarding acceleration of $7.35\, ms^{-2}$ due to frictional force stops the car of mass $400\, kg$ travelling on a road. The coefficient of friction between the tyre of the car and the road is
A$0.55$
B$0.75$
C$6.70$
D$0.65$
Solution
$\begin{array}{l}
As\,we\,know,coefficient\,of\,friction\,\mu = \frac{F}{N}\\
\Rightarrow \mu = \frac{{ma}}{{mg}} = \frac{a}{g}\left( {a = 7.35m{s^{ – 2}}\,given} \right)\\
\therefore \mu = \frac{{7.35}}{{9.8}} = 0.75
\end{array}$
As\,we\,know,coefficient\,of\,friction\,\mu = \frac{F}{N}\\
\Rightarrow \mu = \frac{{ma}}{{mg}} = \frac{a}{g}\left( {a = 7.35m{s^{ – 2}}\,given} \right)\\
\therefore \mu = \frac{{7.35}}{{9.8}} = 0.75
\end{array}$
Standard 11
Physics
