Question

A block of mass $m$ slides down an inclined plane of inclination $\theta$ with uniform speed. The coefficient of friction between the block and the plane is $\mu$. The contact force between the block and the plane is

• A. $\text{mg sin}\theta \sqrt{1+{\mu }^2}$
• B. $\sqrt{(\text{mg sin}\theta {)}^2+(\mu \text{mg cos}\theta {)}^2}$
• C. $\text{mg sin}\theta$
• D. $\text{mg}$

Answer 1

The correct option is D $\text{mg}$

Block slides down with constant velocity. Hence, net force on the block is zero.

So, net contact force $F_C$ on the body is
$F_C=\sqrt{f^2+N^2}$
where $f=\text{mg sin}\theta \{\because a=0\Rightarrow \text{f = mg sin}\theta \}$
and $N=\text{mg cos}\theta$
$\Rightarrow F_C=\sqrt{m^2{g}^2(co{s}^2\theta +si{n}^2\theta )}=\text{mg}$
Hence, the correct answer is (D).