Question

A block of mass 2 kg rests on a rough inclined plane making an angle of 30º 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. 9.8 N
• B. 0.7 × 9.8 × $\sqrt { 3 }$ N
• C. 9.8 × $\sqrt { 3 }$ N
• D. 0.7 × 9.8 N

Answer 1

Answer: (A)

9.8 N

Answer 2

f = mmg cos q = 0.7 × 2 × g × $\frac { \sqrt { 3 } } { 2 }$ = 0.7 × g × $\sqrt { 3 }$

Ext. force = mg sin q = 2 × g × $\frac { 1 } { 2 }$ = 9.8

f > Ext. force. So particle will at rest