A 4.960 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 4.960 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is 0.605 and the coefficient of kinetic friction is 0.305. At time t 0, a force F 18.1 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 Number Number Consider the same situation, but this time the external force Fis 36.5 N. Again state the force of friction on the block at the following times: t> 0 Number Number

Explanation / Answer

Force of static friction = 0.605 × 4.96 × 9.8 =29.41 N

Applied force = 18.1 N

As applied force is less than static friction so force of friction at t = 0 and t > 0 will be same i.e. 18.1N because static friction is self adjusting force.

When applied force = 36.5 N

At t = 0 force of friction = 29.41 N

At t > 0 force of friction will be = 0.305 × 4.96 × 9.8 = 14.83N

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