Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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.)

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.