A 3.030 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 3.030 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?s = 0.605 and the coefficient of kinetic friction is ?k = 0.155. At time t = 0, a force F = 11.1 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 22.3 N. Again state the force of friction acting on the block at the following times: t = 0, t > 0.

Explanation / Answer

a)Fext=11.1N

Ff=.605*3.03*9.81=17.9832N-static

Ff=.155*3.030*9.81=4.607N-kinetic

since Fext<Ffstatic

so no motion

so Ff=F=11.1N for t=0 and t>0

b) F is 22.3 N

Fext>Ff stat

so there is motion

so at t=0 Ff=static=17.9832N

at t>0 Ff=4.607N

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