Work-Energy Theorem Algebraic Representation

[Chrismartin33]

Homework Statement

A car of mass m accelerates from speed v1 to speed v2 while going up a slope that makes an angle θ with the horizontal. The coefficient of static friction is μs, and the acceleration due to gravity is g.
Find the total work W done on the car by the external forces.

Homework Equations

W_net = .5mv₂² - .5mv₁²
W=F*d
W=|F|*|d|*cos(theta)
d=displacement

The Attempt at a Solution

The question actually does say it gives me static friction, not sure how that works since it's a kinetic energy question and must be in motion.
I know the motion is up a ramp, but not quite comprehending how the angle (theta) will affect the various components in this work-energy example.
Answer is algebraic, in terms of any of the following: [v1,v2,(theta),us,g]
Please help!

Edit: I do know that F_net = deltaK/d as well. Not sure if that helps me though.

 

[Chrismartin33]

Going to keep this here incase anyone else gets the answer in the future.
I figured out the answer to the question.
The angle has no effect on the question since velocity is already known, it's simply measuring the change in kinetic energy at the top of the ramp minus the kinetic energy at the bottom of the ramp, and this therefore makes friction an unnecessary portion to the question.
Therefore W_net=deltaK=.5mv₂² - .5mv₁²

 

[Vespa71]

Presuming there's no wind-resistance involved in what is described as "static friction", friction-overcoming work is proportional to distance travelled. So work per meter traveled is constant, and is a sum of work to increase speed, elevate mass, and to overcome friction. Then some equation should be achievable, based on distance traveled as a function of acceleration. If wind resistance is involved, then there's some exponential stuff happening. Which complicate things, but may be solved also. Your figuring is wrong, anyway, as friction and elevation affects how much work is required.

 

[ehild]

Your figuring is wrong, anyway, as friction and elevation affects how much work is required.

@Vespa: Read the OP. Find the total work W done on the car by the external forces:
It was not "work required by the engine".
And recall Work-Energy Theorem: The change of kinetic energy is equal to the total work done. The OP's solution is correct.

 

[Vespa71]

Slapped in the face by definitions, once again. I don't like trick questions too much. A question making my answer correct, would have been a considerably better question. Apologies to Chrismartin33 for my conclusive descriptions. And thanks to ehild for explaining the question in terms that made my journey towards enlightenment possible..

... Humbly Vespa71

 

[deltav]

Thanks Chris, makes sense since the velocity is already accounted for-- my original expression had theta and \mu_s