A 2.590 kg block of wood rests on a steel desk. The coefficient of static fricti

Question

A 2.590 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is s = 0.605 and the coefficient of kinetic friction is k = 0.255. At time t = 0, a force F = 9.45 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times:

a. t=0

b. t>0

Consider the same situation, but this time the external force F is 19.1 N. Again state the force of friction acting on the block at the following times:

c. t=0

d. t>0

Explanation / Answer

Part A)

Ff = uFn = .605(2.590)(9.8) = 15.4 N

Since the required force to move it is greater than the force applied, the frictional force is equal to the applied force

Thus the force is 9.45 N

Part B)

Ff = uFn = .255(2.590)(9.8) = 6.47 N

Parts C and D can use the calculations from parts A and B above since the friction is lower than the applied force.

So Part C = 15.4 N

And Part D = 6.47 N

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