A 2.860 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 2.860 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?s = 0.505 and the coefficient of kinetic friction is ?k = 0.305. At time t = 0, a force F = 8.71 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 17.6 N. Again state the force of friction acting on the block at the following times: t = 0 t > 0

Explanation / Answer

static frcition = 2.86 *10*.505 = 14.4 so initialy the block does not mov friction force actiing = 14.4 2)now that 17.6 is more than static friction so kinetic will come in to picture value = .305 *10*2.86=8.72

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