A 5.790 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 5.790 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 = 0.305. At time t = 0, a force F = 22.9 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 46.1 N. Again state the force of friction acting on the block at the following times: t=0, t>0

Explanation / Answer

The 22.9 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.
Ff = 0.655 * 5.790 * 9.8 = 37.73631 N

At t > 0, the block is moving, so you use the coefficient of kinetic friction.
Ff = 0.305 * 5.790 * 9.8 = 17.36461 N

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