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Jusaquikie
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Block A in the figure below has mass mA = 5.0 kg and is sliding down the ramp. Block B has mass mB = 1.8 kg. The coefficient of kinetic friction between block B and the horizontal plane is 0.50. The inclined plane is frictionless and at angle 30°.
 one year ago
 one year ago
Jusaquikie Group Title
Block A in the figure below has mass mA = 5.0 kg and is sliding down the ramp. Block B has mass mB = 1.8 kg. The coefficient of kinetic friction between block B and the horizontal plane is 0.50. The inclined plane is frictionless and at angle 30°.
 one year ago
 one year ago

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vannayen Group TitleBest ResponseYou've already chosen the best response.1
find what?
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
Equations i have: (x1)FtFk=m1a FT=m1a+(.18*29.4N) FT=(3*3.08) + 5.29=14.5 (x2) m2gsin(30)FT=m2a (x2) 39.2m1a+(.18*29.4N) =m2a (x2) 39.25.29=a(m1+m2) =11a 33.9/11=a=3.08 (y1)FNm1g=0 FN=m1g=29.4N for some reason whenever i come here and write out my work i always see my mistake. maybe i should just type out every problem and save my self the time writing it out 3 times.
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
lol i even paseted the wrong problem
 one year ago

vannayen Group TitleBest ResponseYou've already chosen the best response.1
dw:1348881500104:dw (mB) : (ox) : T  f =mB*a \[f =\mu* mB*g\] we get T  0.5*mB*g = mB*a (1) (mA) :(ox) \[T + mA*g \sin \theta =mA*a (2)\] take (1) + (2) a(mA +mB) = 0.5*mB*g +mA*g*sin30 a = 2.3m/s^2
 one year ago
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