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kengineering
Group Title
starting from rest a 15kg on a horizontal surface acquires kinetic enery due to a constant horizontal force 3ooN . find the kinetic energy acquired after the object has travelled 6m if the coeefficient of friction is 0.10?
 one year ago
 one year ago
kengineering Group Title
starting from rest a 15kg on a horizontal surface acquires kinetic enery due to a constant horizontal force 3ooN . find the kinetic energy acquired after the object has travelled 6m if the coeefficient of friction is 0.10?
 one year ago
 one year ago

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kengineering Group TitleBest ResponseYou've already chosen the best response.0
dw:1349591486655:dw
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
@kengineering What's the force due to friction?
 one year ago

kengineering Group TitleBest ResponseYou've already chosen the best response.0
f14.7
 one year ago

rahulchatterjee Group TitleBest ResponseYou've already chosen the best response.0
Ff=m*a 300(0.10*15*9.8)=15*a => a= 19.02 m/s2 \[v^2=u^2+(2*a*s)\] here u=0.. v=sqrt{2*a*s} => v= 15.10 m/s Kinetic energy= \[1/2 * m*v^2\] = 1711.8 J
 one year ago

Fellowroot Group TitleBest ResponseYou've already chosen the best response.0
is it static friction or kinetic friction?
 one year ago

kengineering Group TitleBest ResponseYou've already chosen the best response.0
kinetic friction
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
F= 300N  u*mg = 285.29N \[W= \int\limits_{0}^{6} 285.29dx =\Delta E = \frac{ 1 }{ 2 }mV ^{2}\] \[ 285.29*6 = \frac{ 1 }{ 2 }mV ^{2}\]
 one year ago
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