A 2.530 kg block of wood rests on a steel desk. Please show all work A 2.530 kg

Question

A 2.530 kg block of wood rests on a steel desk. Please show all work

A 2.530 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is mu s = 0.605 and the coefficient of kinetic friction is mu k = 0.305. At time t = 0, a force F = 9.23 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times: Consider the same situation, but this time the external force . Again state the force of friction acting on the block at the following times:

Explanation / Answer

case 1

maximum static friction force applied can be = 2.53*10*0.605 = 15.306 newton

which means the block will not move untill a force greater than 15.306 N is applied on block

and the frictional force applied by table on block = 9.23 newton

case 2

now the force applied is 18.6 nweton > 15.306

hence block will move

and the frictional force applied by table on block = 2.53*(10)*.305 = 7.716 newton

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