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ksaimouli
 3 years 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?
ksaimouli
 3 years 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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349633059599:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1349633142335:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.1what to do after this i am totally confused

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[F_n=mgF_y=12*9.8Fsin(25)\]

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.1i got this but how do u find the F

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[F_f= \mu F_n= 0.25(117.6Fsin(25))=F_x=Fcos(25)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[29.40.25\sin(25)F=Fcos(25)\] \[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349634909926:dwBasic trig

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No acceleration => no net force in xdirection
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