Question

A block of mass $4kg$ rests on an inclined plane. The inclination of the plane is gradually increased. It is found that when the inclination is $3$ by $5(\sin \theta =\frac{3}{5})$, the block just begins to slide down the plane. The coefficient of friction between the block and the plane is

• A. $0.8$
• B. $0.6$
• C. $0.4$
• D. $0.75$

Answer 1

The correct option is D $0.75$

Given, mass of the block $m=4kg$
so, the situation can be drawn as





From the FBD of the block,
$f_l=\mu N=mg\sin \theta \dots \left(i\right)$
$N=mg\cos \theta \dots \left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$
$\mu mg\cos \theta =mg\sin \theta$
$\mu =\tan \theta$
$\sin \theta =\frac{3}{5}$
clearly, base of the triangle is $4\text{units}$
$\tan \theta =\frac{3}{4}$
$\mu =\tan \theta =\frac{3}{4}=0.75$