Question

A body weighing $20kg$ just slides down a rough inclined plane that rises $5$ in $12$. The coefficient of friction is

• A. $0.46$
• B. $4.6$
• C. $0.52$
• D. $0.12$

Answer 1

Answer: (A)

$0.46$

The given situation can be shown as above
As the plane rises $5$ in $12$
$\therefore \sin \theta =\frac{5}{12}$
and $\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-{\left(\frac{5}{12}\right)}^2}$
When it just slides,
$mg\sin \theta =\mu mg\cos \theta$

So, the coefficient of friction,
$\mu =\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{5}{12}\times \frac{12}{\sqrt{119}}=\frac{5}{\sqrt{119}}=0.46$