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## kengineering 3 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

|dw:1349591486655:dw|

2. ash2326

@kengineering What's the force due to friction?

3. kengineering

f14.7

4. rahulchatterjee

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

tnx

6. Fellowroot

is it static friction or kinetic friction?

7. kengineering

kinetic friction

8. Fellowroot

9. Algebraic!

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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