A 2.910 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 2.910 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is ?s = 0.655 and the coefficient of kinetic friction is ?k = 0.155. At time t = 0, a force F = 11.5 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 and t>0.

Consider the same situation, but this time the external force F is 23.2 N. Again state the force of friction acting on the block at the following times:t=0 and t>0

Explanation / Answer

fsmax = ?s * m * g = 0.655 * 2.910 * 9.8 = 18.68 N

fk = ?k * m *g = 0.155 * 2.910 * 9.8 = 4.42

t=0 ====> f = fs,max = 18.68 N

t>0 ====> f = fk = 4.42 N

Consider the same situation, but this time the external force F is 23.2 N. Again state the force of friction acting on the block at the following times:t=0 and t>0

the answers remain same as it is independent of F

t=0 ====> f = fs,max = 18.68 N

t>0 ====> f = fk = 4.42 N

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