How Is Work Calculated in a Friction-Involved Scenario?

[yue_]

Homework Statement

There is a 2450 kg car being pushed by a group of students. The car is pushed a distance of 12 m in 8.95 seconds. The coefficient of friction is .450.
a) What is the net work being done?
b) Work done by street?
c) Work done by students?

Homework Equations

Wnet=Kf-K0
Wf=-Ff*d
d=[(v0+vf)t]/2
KE=[mv^2]/2

The Attempt at a Solution

For a, Wnet=Kf-K0, and K0=0.
To find Kf, I started to find Vf, using the displacement formula
12=[vf*8.95]/2
vf=2.68 sec.
Then I found Kf
Kf=[2450*2.68^2]/2
Since K0=0, Wnet=8798 J (same value as Kf+0)

b asks for work done by street so I assumed that meant Wf.
At this point, I wasn't sure how to find Ff to use in Wf=-Ff*d formula
I thought Ff=μ*Fn
Is Fn=m*g? Or do I multiply m*a, after finding out acceleration?
Since I found Vf in the last step, do I divide that by time to find acceleration?

c is work done by students, which is Wpush/applied force.
Wnet=Wpull-Wf
Do I add Wf to Wnet to find applied force?
At this point, I don't really know what I'm doing.
Can someone give me a few tips on the concepts that should be understood about this problem and explain whatever errors I'm making?

 

[Zula110100100]

Ff is μ*Fn This will give you the force of friction, and you know the distance over which it was acting. And I am not sure what is better, but I personally would have found the acceleration with s = (at^2)/2
12 = a(8.95^2)/2
24 = 80.1025a
a = .2996m/s^2

Fnet = ma = .2996*2450 = 734.02
Wnet = Fnet*s = 734.02*12 = 8808J

Pretty similar answer

 

[yue_]

Ff is μ*Fn This will give you the force of friction, and you know the distance over which it was acting.

Thanks for replying. I did get that acceleration and a very close Wnet (8820 J) when I worked it previously, while using the method you demonstrated.

For Ff, is Fn=m*a? Or m*g? That's what I'm confused about...

 

[JHamm]

Fn is the normal force, which is mg :)