Question
A Block rests on a horizontal surface.The normal force is 20 N.The coefficient of static friction between the block and the surface is 0,40 and the coefficient of dynamic friction is 0,20.
What Is the magnitude of the frictional force exerted on the block while the block is at rest?
What Is the magnitude of the frictional force exerted on the block while the block is at rest?
Answers
Henry
Fs = u*Fn = 0.4 * 20 = 8 N.