(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Work to push a sled up a hill: Friction, inclines,force,work

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