A 1.890 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 1.890 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is . 0.655 and the coefficient of kinetic friction is = 0.155. At time t 0, a force F = 7.47 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>o Number Number Consider the same situation, but this time the external force Fis 15.1 N. Again state the force of friction acting on the block at the following times: t=0 t>o Number Number

Explanation / Answer

Ffs = µs*m*g

Ffs = 0.655 x 1.890 x 9.81

Ffs = 12.1 N

The 7.47 N horizontal force does not affect the friction force. At t = 0, the block is not moving, so you use the coefficient of static friction

A)

Ff = 7.47 N at all times (applied force < static friction force)

B)

T = 0: Ff = µs*m*g = 12.1 N (The block hasn't moved yet; static friction applies)

T > 0: Ff = µk*m*g = 2.87 N (The block is moving; kinetic friction applies)

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