# Homework Help: Work Done with Friction

1. Feb 28, 2012

### PeachBanana

1. The problem statement, all variables and given/known data

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.

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

3. 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

Last edited: Feb 28, 2012
2. Feb 28, 2012

### 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!

3. Feb 28, 2012

### PeachBanana

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

4. Feb 28, 2012

You bet.