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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?
 2 years ago
 2 years 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?
 2 years ago
 2 years ago

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

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

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

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

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