A 2.380 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 2.380 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 = 9.41 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 19.0 N. Again state the force of friction acting on the block at the following times:

t=0, t>0

Explanation / Answer

The 9.41N 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 * 2.38 * 9.8 = 15.2772 N

At t > 0, the block is moving, so you use the coefficient of kinetic friction.

Ff = 0.305 * 2.38 * 9.8 = 7.1138 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.