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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
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|dw:1349633142335:dw|
what to do after this i am totally confused

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Other answers:

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

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