Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

In a tractor-pull competition, a tractor applies a force of 1.3kN to the sled, which has mass 1.1x 10^4 kg. At that point, the coefficient of kinetic friction between the sled and the ground has increased to 0.80. What is the acceleration of the sled? What is the significance of the sign of the acceleration?

See more answers at
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.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

|dw:1328974050914:dw| i get right along + so \[F-f=ma \] \[f=\mu _{k}mg\] \[a=(F-\mu _{k}mg)/m\]
Im sorry, I dont understand what you wrote down for the formulas Is that f= Mu - overall mass? and what does the other one say?
you can't understand formula ok?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

when F exerted to body, friction so exerted force on it (here not equal)
Are they not equal by units? :S So far what I've done is calculated F(g), which was 1.1x10^4 X 9.81m/s^2, which got me 107910 for Fg Then because FN is assumed to be equal to Fg, I plugged in FN to Ff, which got me 107910 x .80 which got me 86328 I then did FNet= Fapplied - Ffriction, which got me -85028 Plugging that into F=ma, whiuch got me -85028= 1.1x 10^4 (x) Which yielded -7.73 m/s THe answer in the textbook is 0.61m/s Could you tell me what Im doing wrong in my steps? Im sorry for not understanding what you wrote because I still only know the very basics :(
M is overal minus a means that slid going to stop but i distracted ! my solve equal yours !

Not the answer you are looking for?

Search for more explanations.

Ask your own question