Accident Reconstruction using Friction, Energy, Momentum, and Work

[sparris]

(I believe I got this right, but it's necessary context to the next problem.)
PROBLEM 1
1. Homework Statement
"The police department determined that the force required to drag a 130 N (29 lb) car tire across the pavement at a constant velocity is 100 N (23 lb). Specifications from the truck’s manufacturer claim that (for technical reasons) the effective coefficient of friction for truck tires is only 70% that of car tires. What was the coefficient of kinetic friction between the tires and the road for both the car and truck?"

since F= ma and there's a constant velocity || ƩF_x = 0 N
F_a = 100 N

F_g = 130 N

I'm also given that the angles traveled after the accident are 7° for the truck and 33° for the car.2. Homework Equations
μ = F_f/F_n

3. The Attempt at a Solution

ƩF_y = F_g - F_n = 0
F_n = 130 N

ƩF_x = F_a - F_f = F_a - μ_cF_n
0 = 100 N - μ(130 N)
(of car) μ_c = 10/13 = 0.769

7μc/10 = μ_t
(of truck) μ_t = 70/130 = 0.538

PROBLEM 2
1. Homework Statement
After collision, the truck and car skidded at the angles shown in the diagram. The car skidded a distance of 8.2 m (27 ft) before stopping while the truck skidded 11 m (37 ft) before stopping. The weight of the car is 13,600 N (3050 lb) and the weight of the truck is 69,700 N (15,695 lb). What was the speed of each vehicle just after the collision?

μ_c = 0.769
μ_t = 0.538
ΔD_c = 8.2 m
ΔD_t = 11 m
Fg_c = 13,600 N
Fg_t = 69,700 N

2. Homework Equations
μ = Ff/Fn
W_f = F_fΔD
W_f = μF_gΔD = ΔK = 1/2mv²
μF_gΔD = 1/2mv²
μgΔD = 1/2v²
v = sqrt(2μgΔD)

3. The Attempt at a Solution

(for car) v = sqrt(2μ_cgΔD_c)
v_c = sqrt(2(0.769)(9.81)(8.2)) = 11.1 m/s

(for truck) v = sqrt(2μ_tgΔD_t)
v_t = sqrt(2(0.538)(9.81)(11)) = 10.8 m/s

I don't think these are the correct answers, but I'm not sure where the flaw in my solving lies. Help, please?
Thanks in advance. :)

Link to actual page:
physics.info/momentum-two-three/accident-reconstruction-1.pdf

 

[mfb]

I don't think these are the correct answers

Why? I didn't check all numbers (a computer can do this), but the formulas and the order of magnitude look correct.

 

[sparris]

Perhaps I've just gotten too used to second guessing myself on my calculations. Sorry! Thanks for letting me know though. The fact that he gave me weight and I didn't use it threw me off.