Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

4. A 1.2 kg block of mass is placed on an inclined plane that has a slope of 30° with respect to the horizontal, and the block of mass is released. A friction force Fk = 2.0 N acts on the block as it slides down the incline. Find the acceleration of the block down the incline

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

The free body diagram of the problem has 3 forces: 1) The Normal perpendicular to inclined plane. n=m*g=1.2*10=12 N 2) The friction parallel to inclined plane f=2.0 N 3) The weight that has two components, one perpendicular and other parallel (n*sin(30)) The Newton equation parallel to inclined plane is: \[-f + m \times g \sin (30) = m \times a\] \[a=acceleration=-f \div m + g \times \sin (30)= -2 \div 1.2 + 5 = 3.33 meters/\sec^{2}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question