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7. A stationary 25 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. A horizontal force of 300 N is applied to the object. Determine the force of friction. *

Sagot :

Answer:

  73.5 N opposite the applied force

Explanation:

The normal force the object exerts on the table is ...

  F = ma

  F = (25 kg)(9.8 m/s^2) = 245 N

The force required to overcome the static friction and set the object in motion is ...

  (0.50)(245 N) = 122.5 N

The applied force is more than sufficient for that purpose, so we must assume the object is being accelerated.

The force of friction while the object is in motion is ...

  (0.30)(245 N) = 73.5 N . . . opposing the applied horizontal force

_____

Additional comment

Friction force is in the direction opposite to the direction of motion along a surface. The magnitude of it is the product of the coefficient of (static or kinetic) friction and the normal force between the surface and the object. Usually, the coefficient of static friction is greater than that of kinetic friction. (Once an object is set in motion, it usually takes less force to keep it in motion.)