How Do You Calculate the Coefficient of Friction for a Slowing Truck?

[BrownBoi7]

A 4700 kg truck carrying a 900 kg crate is traveling at 25 m/s to the right along a straight, level road, as shown below.

Suppose the brakes are applied for 3.0s. During this 3.0s the truck travels 55 m.

1. Calculate the acceleration of the truck during the 3.0s interval. (Assume it is constant.)

2. Sketch a force diagram of the crate during the 3.0s interval described above.

3. Calculate the minimum coefficient of friction required to keep the crate from sliding in the back of the truck.

4. Calculate the minimum coefficient of friction required to stop the truck by the end of the 3.0s interval

My attempt:
1. Using X = Xo + Vot - 1/2at^2 a = -4.4 m/s^2
2. I got it.
3. F = ma ; F = uN
uN = ma ==> umg = ma ==> u = a/g = -4.4/9.8 = 0.449
4. This is where I need help. Since it was never mentioned that the truck came to a complete rest with -4.4 m/s^2 in 3 seconds. But now we do want it to stop in 3 secs. So I am guessing negative acceleration has to be greater.
a = (0-25 m/s)/3 sec = -8.33 m/s^2
u = a/g ==> -8.33/9.8 = 0.850
Somebody please review my answers. Did I go about them the right way?

Thanks!

 

[mfb]

Looks good.
The problem statement in (4) is a bit unclear - for example, truck and crate could have different coefficients of friction, or the crate could have the coefficient from (3) and the truck has to compensate for that, or whatever. Well, I think your interpretation is the way the question is meant.