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

In summary, the question asks how much work is done in pushing a 35 kg sled up a 3.6 m hill inclined at 15 degrees with a horizontal push and a coefficient of kinetic friction of 0.2. The normal force is calculated to be 331.313N, the force down the hill is 88.77N, and the friction force is 66.262N. With constant velocity, the forces will equal out, resulting in an uphill push force of 155.038N. Two attempts are made to calculate the work done: 2156J and 2232J. However, the correct answer is 2300J, indicating an error in the calculations.
  • #1
krausr79
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0

Homework Statement


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?


Homework 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

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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  • #2
krausr79 said:
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! :smile:

No, the push is horizontal, so that will add to the usual normal force. :wink:
 
  • #3
Thank you.
 

1. How does friction affect the amount of work required to push a sled up a hill?

Friction is a force that acts in the opposite direction of motion, so it makes pushing a sled up a hill more difficult. The amount of work required to overcome friction depends on the type of surface and the weight of the sled.

2. Does pushing a sled up an incline require more or less force compared to pushing it on a flat surface?

Pushing a sled up an incline requires less force because the force of gravity is partially counteracted by the slope of the hill. This means less force is needed to overcome the force of gravity and friction.

3. How does the angle of the incline affect the amount of work required to push a sled up a hill?

The steeper the incline, the more work is required to push a sled up a hill. This is because the steeper the incline, the greater the component of the sled's weight that needs to be overcome in order to move it up the hill.

4. Is the amount of work done to push a sled up a hill affected by the weight of the sled?

Yes, the amount of work required to push a sled up a hill is directly proportional to the weight of the sled. The heavier the sled, the more work is required to overcome the force of gravity and friction.

5. How is work calculated when pushing a sled up a hill?

Work is calculated by multiplying the force applied to the sled by the distance it is moved. In the case of pushing a sled up a hill, the force applied is the force needed to overcome the force of gravity and friction, and the distance moved is the horizontal distance the sled is pushed up the hill.

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