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

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# Homework Help: Work Done with Friction

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