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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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ksaimouliBest ResponseYou've already chosen the best response.1
dw:1349633142335:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
what to do after this i am totally confused
 one year ago

henpenBest ResponseYou've already chosen the best response.2
\[F_n=mgF_y=12*9.8Fsin(25)\]
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
i got this but how do u find the F
 one year ago

henpenBest ResponseYou've already chosen the best response.2
\[F_f= \mu F_n= 0.25(117.6Fsin(25))=F_x=Fcos(25)\]
 one year ago

henpenBest ResponseYou've already chosen the best response.2
\[29.40.25\sin(25)F=Fcos(25)\] \[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]
 one year ago

henpenBest ResponseYou've already chosen the best response.2
dw:1349634909926:dwBasic trig
 one year ago

henpenBest ResponseYou've already chosen the best response.2
No acceleration => no net force in xdirection
 one year ago
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