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

by krausr79
Tags: force, friction, hill, inclines, push, sled, work
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krausr79
#1
Oct3-08, 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?
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tiny-tim
#2
Oct4-08, 03:43 PM
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Quote Quote by krausr79 View Post
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?

Normal force = 35*9.8*cos(15) = 331.313N
Hi krausr79!

No, the push is horizontal, so that will add to the usual normal force.
krausr79
#3
Oct5-08, 11:16 AM
P: 40
Thank you.


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