Running Out of Gas - work & power problem

[DraculaNotI]

1. Homework Statement :
You are at the top of a hill that is 150 m in height and are traveling at 36 km/h when your car runs out of gas. Due to the relatively shallow grade, the road down the hill is 3000 m long. After rolling down the hill, the car continues to roll on a horizontal section of road. The mass of the car and driver is 1000 kg and on both the hill and the horizontal stretch, the rolling friction is equal to 400 N. Neglect air drag in this problem and use g = 10 m/s².


2. Homework Equations :
Assuming you do not apply the brakes, what is your speed v_f when you arrive at the bottom of the hill? Remember that this answer will be an overestimate since we are neglecting air drag.
Give your answer to 3 significant figures in km/h. Pay careful attention to your signs in this question!


3. The Attempt at a Solution :
I've tried been using these equations that we went over in class...
K_f - K_i = -f*d (not sure if this is relevant)
W_friction = \DeltaU + K_f - K_i
Converting v_i to m/s: 36*1000/3600 = 1
(400*3000) = 1,500,000 + (1/2)(1000)*v_f² - (1/2)(1000)(1)
...which ends up in a disaster of imaginary numberness if I'm punching it all in correctly. D:
I've also tried some other methods (putting minus signs wherever), but my answers (I've gotten 176m/s and 88.3m/s) have come up as wrong every time.

Any help would be greatly appreciated! :)

 

[willem2]

W_friction = \DeltaU + K_f - K_i

Converting v_i to m/s: 36*1000/3600 = 1
(400*3000) = 1,500,000 + (1/2)(1000)*v_f² - (1/2)(1000)(1)
...which ends up in a disaster of imaginary numberness if I'm punching it all in correctly. D:

36*1000/3600 = 10, not 1

The work done by friction is always negative, and so is \Delta U because the
car has less gravitational potential energy when it has moves down the hill.

 

[DraculaNotI]

Thank you, willem2! It makes a lot more sense now (plus I've had a good 9 hours of sleep). :)

Lesson learned: don't try to convert units in your head at 1am