A 5.430 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 5.430 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is mu s= 0.555 and the coefficient of kinetic friction is mu s =0.255. At time t= 0, a force F = 18.2 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 Number Number Consider the same situation, but this time the external force F is 36.7 N Again state the force of friction acting on the block at the following times: t = 0 t >0 Number Number

Explanation / Answer

when F = 18.2 < (0.555*9.8*5.43 = 29.5337 )

then Fric static force = F = 18.2 N

second case

F = 36.7 >(0.555*9.8*5.43 = 29.5337 )

fric force = 0.255*9.8*5.430 =13.569 N

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