A 2.450 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 2.450 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 = 0.255. At time t = 0, a force F = 7.46 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 15.1 N. Again state the force of friction acting on the block at the following times: t = 0 and t > 0

Explanation / Answer

Ffs = µs*m*g

Ffs = 2.45 * 9.8 * 0.505 = 12.125 N

A) Ff = 7.46 N at all times (applied force < static friction force)

B)
T = 0: Ff = µs*m*g = 12.125 N (The block hasn't moved yet; static friction applies)
T > 0: Ff = µk*m*g = 6.125 N (The block is moving; kinetic friction applies)

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