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kengineering

  • 2 years ago

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?

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  1. kengineering
    • 2 years ago
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    |dw:1349591486655:dw|

  2. ash2326
    • 2 years ago
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    @kengineering What's the force due to friction?

  3. kengineering
    • 2 years ago
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    f14.7

  4. rahulchatterjee
    • 2 years ago
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    F-f=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

  5. kengineering
    • 2 years ago
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    tnx

  6. Fellowroot
    • 2 years ago
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    is it static friction or kinetic friction?

  7. kengineering
    • 2 years ago
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    kinetic friction

  8. Fellowroot
    • 2 years ago
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  9. Algebraic!
    • 2 years ago
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    F= 300N - u*mg = 285.29N \[W= \int\limits_{0}^{6} 285.29dx =\Delta E = \frac{ 1 }{ 2 }mV ^{2}\] \[ 285.29*6 = \frac{ 1 }{ 2 }mV ^{2}\]

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