# Force of friction on different times

1. Oct 24, 2012

### charlygarcia

1. The problem statement, all variables and given/known data

A 3.280 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.605 and the coefficient of kinetic friction is μk = 0.155. At time t = 0, a force F = 12.0 N is applied horizontally to the block. State the force of friction applied to the block by the table at t=0 and t>0:

Consider the same situation, but this time the external force F is 24.1 N. Again state the force of friction acting on the block at t=0 and t>0:

2. Relevant equations
F = ma
F ≤ μsN
F = μkN
w=mg

3. The attempt at a solution
I tried to find the normal force with mg so I could then multiply this with μs and μk. I thought that the first one would be when time equals 0 but I don't really understand how to use the F = 12(first question) or F = 24.1(second question).
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 24, 2012

### SHISHKABOB

well so what is the static frictional force acting on the block?

What happens to the block if the force that is applied to it is *less than* the static frictional force? What about if the force applied to it is *greater than* the static frictional force?