A 5.290 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 5.290 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?s = 0.505 and the coefficient of kinetic friction is ?k = 0.255. At time t = 0, a force F = 16.1 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times:

1) t=0

2)t>0

Consider the same situation, but this time the external force F is 32.5 N. Again state the force of friction acting on the block at the following times:

1) t=0

2)t>0

Explanation / Answer

max. value of static friction = us .Mg = 0.505 x 5.290 x 9.8 =26.18 N

applied force F = 16.1 N which is less than limit forfriction force.

so block will not move

so friction force = 16.1 N will work for t = 0 and t>0

F =32.5 is greater than limi t (26.18 N)

so at t = 0 frcition   = 26.18 N

at t>0
kinetic friction will work

friction = 0.255 x 5.29 x9.8 =13.22 N

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