A 6.310 kg block of wood rests on a steel desk. The coefficient of static fricti
ID: 1428761 • Letter: A
Question
A 6.310 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is pi = 0.555 and the coefficient of kinetic friction is mu_s = 0.555 and the coefficient of kinetic friction is mu_k = 0.255. At time t= 0, a force F = 21.1 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times: Incorrect. The block is not moving across the surface in this situation. Consider the same situation, but this time the external force F is 42.6 N. Again state the force of friction acting on the block at the following times:Explanation / Answer
static friction = us*N = usmg = 0.555*6.310*9.8 = 34.320 N
kinetic friction = ukmg = 0.255*6.310*9.8 = 15.7686 N
now at F 21.1 N
as it less them Max of static friction so static friction will act and = Fapplied ( to stop the block)
=> friction = 21.1 N at t=0 also friction = 21.1 N at t>0
if F = 42.6 N applied
max of friction will act at t=0 , = 34.320 N and then kinetic friction will act later t>0 = 15.7686 N