
#1
Oct308, 08:55 PM

P: 40

1. The problem statement, all variables and given/known data
We are pushing a sled up a hill. Our push is horizontal. The hill is inclined at 15 degrees. The sled masses 35 kg. The coefficient of kinetic friction is .2. The hills height is 3.6 m. We push at a constant velocity. How much work do we do in pushing? 2. Relevant equations Mass*Gravity Accel*cos(15) = Normal force/Gravity into surface Mass*Gravity Accel*sin(15) = Gravity slideways Normal*Coef of friction = friction force sin(x) = Opp/hyp or hyp = Opp/sin(x) Work = Force*Distance 3. The attempt at a solution Normal force = 35*9.8*cos(15) = 331.313N Force down hill = 35*9.8*sin(15) = 88.77N Friction force = 331.313*.2 = 66.262N Length of incline = 3.6*sin(15) = 13.909M With constant velocity, forces will equal out gravity and friction = uphill push force = 66.262+88.77 = 155.038N Work = 155.038*13.909 = 2156J ? I thought this would be it assuming the force and distance must be in the same direction. forward push*cos(15) = uphill push forward push = Uphill push/cos(15) = 155.038/cos(15) = 160.507 Work = 160.507*13.909 = 2232J ? The back of book answer is 2300J (2 signifigant digits) Where is the error? 



#2
Oct408, 03:43 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

No, the push is horizontal, so that will add to the usual normal force. 



#3
Oct508, 11:16 AM

P: 40

Thank you.



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