A 2.360 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 2.360 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 = 8.61 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 17.4 N. Again state the force of friction acting on the block at the following times:t = 0,t > 0.

Explanation / Answer

Here, Normal reaction (R) = mg

= 23.12 N

=> Maximum Static friction = sR (i.e., coefficient of static friction*Normal reaction)

= 13.98 N

=> Kinetic friction = kR ( i.e., coefficient of Kinetic friction*Normal reaction)

= 3.58 N

For a body to move, applied force should be greater than Maximum static friction then and only a body moves. If the applied force is less than maximum static friction, then frictional force = applied force.

But once set in motion frictional force acting on the body is kinetic friction.

In the first case,

Applied force,F=8.61 N which is less than Maximum static friction of 13.98 N.

So body doesn't move. So at t=0 & t > 0, frictional force = F = 8.61 N

In the Second case,

Applied force,F=17.41 N which is greater than Maximum static friction of 13.98 N.

So, body moves. So at t=0, frictional force = F = 13.98 N

Since the body is set in motion now at t>0, kinetic friction acts.

So, at t>0, frictional force =Kinetic friction = 3.58 N

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