Question

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

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

Answer 1

The correct option is A $9.8N$

The force applied on the body that is on the inclined plane is given as,

$F=mg\sin \theta$

$F=2\times 9.8\times \sin {30}^{\circ }$

$=9.8N$

The limiting friction force between the block and the inclined plane is given as,

$f=\mu mg\cos \theta$

$f=0.7\times 2\times 9.8\cos {30}^{\circ }$

$=11.88N$

Since the limiting friction force is greater than the force that tends to slide the body.

Thus, the body will be at rest and the force of friction on the block is $9.8N$.