Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Josh1012

  • 2 years ago

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

  • This Question is Closed
  1. galanh
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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: \[-f + m \times g \sin (30) = m \times a\] \[a=acceleration=-f \div m + g \times \sin (30)= -2 \div 1.2 + 5 = 3.33 meters/\sec^{2}\]

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.