On a steel desk rests a 2.420-kg block of wood. The coefficient of static fricti
ID: 1279122 • Letter: O
Question
On a steel desk rests a 2.420-kg block of wood. The coefficient of static friction between the desk and the block is ?s = 0.455 and the coefficient of kinetic friction is ?k = 0.205. At time t = 0, a force F = 6.64 N is applied horizontally to the block. What is the force of friction applied to the block by the table at the following times?
t=0 and t>0
Consider the same situation, but this time the external force F is 13.4 N. Again find the force of friction acting on the block at the following times:
t=0 and t>0
Explanation / Answer
The weight of the block = 2.420 * 9.8 = 23.72 N
Force of Static friction = 23.72 * 0.455 = 10.79 N
The force of kinetic friction which is just the Weight times the coefficient
= 23.72 * 0.205 = 4.86