What is the minimum work required to push a car up an incline with friction?

[PeachBanana]

Homework Statement

What is the minimum work needed to push a 900 kg car 880 m up along a 9.5° incline? Assume the effective coefficient of friction retarding the car is 0.20.

Homework Equations

W = F * d * cos (θ)
Normal Force - mg * cos (9.5°)
Force of kinetic friction = μ * Normal Force

The Attempt at a Solution

I drew out a free body diagram. The first thing I did:

Normal Force - y component of gravity = 0 because it is not accelerating in the y direction.
Normal Force - mg * cos (9.5°).
Normal Force = mg * cos (9.5°).
Normal Force = (900 kg)(9.8 m/s^2) * cos (9.5°)
Normal Force = 1.73*10^3 N

Second: Force of Kinetic Friction = (0.20)(8.69*10^3 N)
Force of Kinetic Friction = 1.73*10 ^3N

Work done by Friction : (1.73*10^3)(880 m) * sin 9.5°
W = 2.5*10^5 J

2.5*10^5 J + 1.28 *10^6 J (This is the minimum work required without friction. The first part of the question asked this and it confirmed this was the correct answer) =

1.53 * 10^6 J

 

[Delphi51]

I get 1.74 x 10^3 rather than your 1.73 for the friction force; don't know if that is important.

In finding the work, why multiply by sin (9.5) ? The force of friction and the distance are in the same direction so no sine factor should be used.

The rest of your calc looks good!

 

[PeachBanana]

Oops! I forgot - friction acts over the entire distance, not just the height as I was thinking.

 

[Delphi51]

You bet.

 

[asteriatic]

The minimum work needed to push the car up the incline with friction is 1.53 * 10^6 J. This takes into account the work done by friction, which is equal to the force of kinetic friction multiplied by the distance traveled up the incline, and taking into account the angle of the incline. The normal force, which is needed to calculate the force of kinetic friction, is also taken into account by considering the y component of gravity. It is important to note that this is the minimum work required, as friction will always oppose motion and therefore require additional work to be done.