A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
can the acceleration of two bodies wth different masses be the same provided they are each release on the same plane with the same angle of inclination one after the other?
anonymous
 4 years ago
can the acceleration of two bodies wth different masses be the same provided they are each release on the same plane with the same angle of inclination one after the other?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes they both can be the same if they are released with different forces F=m*a 2=1*a a=2 4=2*a a=2 i used different masses and still found the same acceleration. Do you understand now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep but you need to resolve the forces so i taugth it will be a=(gcos n friction)/mass where n is the angle made by the plane with the horizontal .so the acceleration depends on mass

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i think you made an error its not linear motion but on inclined plane

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i haven learned this yet..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0excuse accepted but i think you need to answer questions of your level in other not to give false answers

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As an example consider the situation depicted . consider a freebody diagram shows the forces acting upon a 100kg crate that is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate. Begin the above problem by finding the force of gravity acting upon the crate and the components of this force parallel and perpendicular to the incline. The force of gravity is 980 N and the components of this force are Fparallel = 490 N (980 N • sin 30 degrees) and Fperpendicular = 849 N (980 N • cos30 degrees). Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight vector). The force of friction can be determined from the value of the normal force and the coefficient of friction; Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 • 849 N). The net force is the vector sum of all the forces. The forces directed perpendicular to the incline balance; the forces directed parallel to the incline do not balance. The net force is 235 N (490 N  255 N). The acceleration is 2.35 m/s/s (Fnet/m = 235 N/100 kg).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think may be this is the answer to my question can some body verify to tell me it true
Ask your own question
Sign UpFind 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
 Engagement 19 Mad Hatter
 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.