Question

A block $B$ of mass $2kg$ rests on a rough inclined plane making an angle of ${30}^o$ with the horizontal surface. The coefficient of static friction between the block and the plane is $0.7$. The friction force on the block is:

• A. $9.8N$
• B. $0.7\times 9.8\sqrt{3}N$
• C. $9.8\times 7N$
• D. $0.8\times 9.8N$

Answer 1

The correct option is B $0.7\times 9.8\sqrt{3}N$

Given that,

Mass $m=2kg$

Angle $\alpha ={30}^0$

Friction coefficient $\mu =0.7$

Now, according to diagram

Normal force $N=mgcos\alpha$

Friction$=\mu N$

Now, the friction force

$F_f=\mu N$

$F_f=0.7\times 2\times 9.8\times \frac{\sqrt{3}}{2}$

$F_f=0.7\times 9.8\times \sqrt{3}N$

Hence, the friction force is $0.7\times 9.8\times \sqrt{3}N$