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?
Fs = u*Fn = 0.4 * 20 = 8 N.
To find the magnitude of the frictional force exerted on the block while it is at rest, we need to use the equation for frictional force, which is given by:
Frictional force = Coefficient of static friction * Normal force
Given:
Coefficient of static friction = 0.40
Normal force = 20 N
Plugging the values into the equation:
Frictional force = 0.40 * 20
Frictional force = 8 N
Therefore, the magnitude of the frictional force exerted on the block while it is at rest is 8 N.