A body of mass 2 kg rests on a rough inclined plane making an angle of 30^(@) 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 body of mass 2 kg rests on a rough inclined plane making an angle of

1.	A body of mass 2 kg rests on a rough inclined plane making an angle of 30^(@) with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is :
Answer» 9.8 N `9.8 sqrt3N` `0.7 xx 9.8 N` `0.7 xx 9.8 xx 3N` Solution :Here FORCE of limiting friction`F_(f)=muR=mu mg cos 30^(@)` `=0.7xx2xx9.8xx(sqrt3)/(2)\=11.88N` Downward force responsible for MOTION `=mg sin 30^(@) = 2 xx 9.8 xx(1)/(2)=9,8N` Since mg sin is less than `F_(f)` the block does not move. THUS force of friction = force down the INCLINED plane `:. F_(f)=9.8 N` Hence correct choice is (a).

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A body of mass 2 kg rests on a rough inclined plane making an angle of 30^(@) with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is :

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