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 || 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

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mfb

Mentor
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.

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