Question

A block of mass $3\text{kg}$ slides on a rough fixed inclined plane of ${37}^{\circ }$angle , having coeffecient of friction $0.5$. The resultant force exerted by plane on the block is -

• A. $6\sqrt{5}\text{N}$
• B. $8\sqrt{5}\text{N}$
• C. $10\sqrt{5}\text{N}$
• D. $12\sqrt{5}\text{N}$

Answer 1

The correct option is D $12\sqrt{5}\text{N}$

FBD of block:


The plane can exert two forces on the block (i) Friction (ii) Normal reaction force.

If angle of inclination is greater than angle of repose, friction is kinetic in nature , $\tan \varphi \le \mu$

So, $f_k=\mu N=\mu mg\cos {37}^{\circ }$

$f_k=0.5\times 3\times 10\times \frac{4}{5}=12\text{N}$ .........(i)

And,

$N=mg\cos {37}^{\circ }=3\times 10\times \frac{4}{5}=24\text{N}$.........(ii)

$\therefore$ Net force on block by the plane is,

$F=\sqrt{(f_k{)}^2+N^2}=\sqrt{{12}^2+{24}^2}$

$F=12\sqrt{5}\text{N}$

Hence, option (d) is the correct answer.

Why this question? Key concept: Angle of repose is maximum angle of inclination such that a block placed on a inclined plane remain at rest.