# A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity. (a) What is the coefficient of sliding friction between the block and the tabletop? (b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity?

**Friction Numerical Problem) **

A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity.

(a) What is the coefficient of sliding friction between the block and the tabletop?

(b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity?

**Solution**

a) Weight of the block = W = mg = 40N

In this case the normal force = F_{N} = W = 40N

Let the coefficient of sliding friction between the block and the tabletop = μ

Then, sliding Friction between the block and the tabletop = F_{F}= μF_{N} = μ 40 = 40μ

As per the question, a force of 14 N is just required to keep the block moving at a constant speed.

Hence, sliding friction magnitude = applied force magnitude

40μ = 14

μ = 14/40 = 0.35

b) In the second case, the normal force = F_{N} = W = (40 + 20) N= 60 N

Normal force = F_{N} = W = 60N

Friction = F

_{F}= μF_{N}= 0.35 x 60 =21 N

Hence this time to move at a constant velocity the applied force has to be equal to 21 N