ksaimouli
  • ksaimouli
A block is pulled by a string that makes an angle of 25º to the horizontal. If the mass of the block is 12.0 kg and the coefficient of friction is 0.25, what force would keep the block moving at a constant velocity?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1349633059599:dw|
ksaimouli
  • ksaimouli
|dw:1349633142335:dw|
ksaimouli
  • ksaimouli
what to do after this i am totally confused

Looking for something else?

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

More answers

ksaimouli
  • ksaimouli
i know that ff=u*fn
ksaimouli
  • ksaimouli
we need to find fn
anonymous
  • anonymous
\[F_n=mg-F_y=12*9.8-Fsin(25)\]
ksaimouli
  • ksaimouli
i got this but how do u find the F
anonymous
  • anonymous
\[F_f= \mu F_n= 0.25(117.6-Fsin(25))=F_x=Fcos(25)\]
ksaimouli
  • ksaimouli
how did u get 117.6
anonymous
  • anonymous
mg
anonymous
  • anonymous
12*9.8
anonymous
  • anonymous
\[29.4-0.25\sin(25)F=Fcos(25)\] \[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]
ksaimouli
  • ksaimouli
thx
ksaimouli
  • ksaimouli
why =fcos(25)
anonymous
  • anonymous
|dw:1349634909926:dw|Basic trig
anonymous
  • anonymous
\[F_f=F_x=Fcos(25)\]
anonymous
  • anonymous
No acceleration => no net force in x-direction

Looking for something else?

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