A 6.040 kg block of wood rests on a steel desk. The coefficient of static fricti
ID: 1365086 • Letter: A
Question
A 6.040 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is s = 0.655 and the coefficient of kinetic friction is k = 0.205. At time t = 0, a force F = 23.9 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times:
t=0
___________ N
t >0
___________N
Consider the same situation, but this time the external force F is 48.1 N. Again state the force of friction acting on the block at the following times:
t = 0
_________N
t > 0
_________N
Explanation / Answer
for the applied force , F = 23.9 N
at t = 0
as there is no applied force , frictional force = 0 N
at t > 0
maximum static friction = us * m* g
maximum static friction = 0.655 * 6.040 * 9.8
maximum static friction = 38.8 N
as applied force is less than the maximum static friction , the block will not move
frictional force = 23.9 N
--------------------------------
for applied force = 48.1 N
at t = 0
as there is no applied force , frictional force = 0 N
for t > 0
as applied force is greater than maximum static friction
the block will move and
frictional force = uk * m * g
frictional force = 0.205 * 9.8 * 6.040
frictional force =12.13 N
the frictional force is 12.13 N