Question

a block of mass 20kg is placed on a rough horizontal surface when a horizontal force of 200N is applied on the body it just begins to move then the angle of friction is (g = 10m/s^2)
a) 30°
b) 60°
c) 45°
d)90°​

Answer 1

Given:-

• Force= 200N
• Mass of block= 20 kg
• g= 10 m/s²

To Find:-

• Angle of friction

Solution:-

as we know the Force applied on the block is On Horizontal Direction which is 200N

and

Norma(mg)l is Acting downward which is= 20kgx10m/s²= 200N

So,

the Angle of friction is the angle between Normal and The Frictional force applied on the block

which is 90°

$\therefore \: the \: angle \: of \: friction = 90\degree$