Work of a car's engine as it moves up a slope

[v0rtexza]

Homework Statement

A car of mass 900 kg accelerates up a slope. The velocity of the car at the bottom of the slope is 5 m/s. By the time the car reaches the top of the slope, its velocity is 15 m/s. The slope is 8m high from the ground and has a length (radius - angled length - not base) of 500m. The average frictional force experienced by the car between A and B is 50N.

Calculate: The total work done by the car's engine between the bottom and top of the slope.

Homework Equations

F = ma
m = mass in kilograms
a = acceleration in m/s
F = force in Newtons
W = F x s
s = distance
W = work

The Attempt at a Solution

I am quite at a loss with this question but this is my attempt.
Fbottom = m x a = 900 x 5 = 4500N

Work (bottom) = f x s
= 4500 x 500 (slope is 500m)
= 2250000J

Ftop = F x ma
= 900 x 15
= 13500N
Work (top) = f x s
= 135000 x 500
= 6750000J

W (top) - W (bottom) = 450000J

I am sure that my answer is incorrect but I am unsure as to how to solve it and the memo is missing the answer, any help would be appreciated. Sorry for the n00b question! :)

 

[CAF123]

its acceleration is 15 m/s

There is a typo here somewhere, you mean its velocity is 15m/s?

 

[v0rtexza]

Yeah, sorry about that!

 

[CAF123]

It is difficult to follow your solution because you have inputted velocities in place of accelerations.
I think this is meant to be solved using the work-energy theorem, but also incorporating the effects of an external force, ie friction, so; $$W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 + F_f d = ΔK.E + F_f d,$$ where $F_f$ is the force of friction.

 

[CWatters]

Don't forget the gain in PE as well as KE?

 

[CAF123]

In taking into account the change in potential energy, i believe we add on, to the eqn above;
$$W = \vec{F}.\vec{d} = mgdcos(90 +θ) = -mgdsinθ$$

Apologies for forgetting that earlier, I must have done the problem ignoring the incline!