Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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 :

sqdancefan

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.)

We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.