A 5.950 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 5.950 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is s = 0.455 and the coefficient of kinetic friction is k = 0.305. At time t = 0, a force F = 16.3 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,t>0

Consider the same situation, but this time the external force F is 32.9 N. Again state the force of friction acting on the block at the following times:

t=0,t>0

Explanation / Answer

m = 5.950 Kg
us = 0.455
uk = 0.305

(a)
F = 16.3 N

At t = 0, Static Friction will be in action
Friction Force at rest = us * m*g
Fs = 0.455 * 5.950 * 9.8 N
Fs = 26.53 N

At t > 0
As the Applied Force < Friction Force block would not move and therefore Friction Force, Fs = 26.53 N

(b)
F = 32.9 N
At t = 0 ,
Same as above Static Friction will be in action, & therefore Fs = 26.53 N

At t > 0
As Applied Force, > Friction Force , The block will start moving and so Kinetic Friction will be in action,
Fk = uk * m*g
Fk = 0.305 * 5.950 * 9.8 N
Fk = 17.78 N

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