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Accident Reconstruction using Friction, Energy, Momentum, and Work

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(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 || ƩFx = 0 N
Fa = 100 N

Fg = 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 = Ff/Fn

3. The Attempt at a Solution

ƩFy = Fg - Fn = 0
Fn = 130 N

ƩFx = Fa - Ff = Fa - μcFn
0 = 100 N - μ(130 N)
(of car) μc = 10/13 = 0.769

7μc/10 = μt
(of truck) μt = 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?

μc = 0.769
μt = 0.538
ΔDc = 8.2 m
ΔDt = 11 m
Fgc = 13,600 N
Fgt = 69,700 N

2. Homework Equations
μ = Ff/Fn
Wf = FfΔD
Wf = μFgΔD = ΔK = 1/2mv2
μFgΔD = 1/2mv2
μgΔD = 1/2v2
v = sqrt(2μgΔD)

3. The Attempt at a Solution

(for car) v = sqrt(2μcgΔDc)
vc = sqrt(2(0.769)(9.81)(8.2)) = 11.1 m/s

(for truck) v = sqrt(2μtgΔDt)
vt = 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
 
Last edited:
33,372
9,098
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.
 
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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.
 
Last edited:

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