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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 || ƩF

F

F

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

3. The Attempt at a Solution

ƩF

F

ƩF

0 = 100 N - μ(130 N)

(of car) μ

7μc/10 = μ

(of truck) μ

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

μ

ΔD

ΔD

Fg

Fg

2. Homework Equations

μ = Ff/Fn

W

W

μF

μgΔD = 1/2v

v = sqrt(2μgΔD)

3. The Attempt at a Solution

(for car) v = sqrt(2μ

v

(for truck) v = sqrt(2μ

v

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

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 NF

_{a}= 100 NF

_{g}= 130 NI'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}= 0F

_{n}= 130 NƩF

_{x}= F_{a}- F_{f}= F_{a}- μ_{c}F_{n}0 = 100 N - μ(130 N)

(of car) μ

_{c}= 10/13 = 0.7697μc/10 = μ

_{t}(of truck) μ

_{t}= 70/130 = 0.538PROBLEM 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}= 0.769μ

_{t}= 0.538ΔD

_{c}= 8.2 mΔD

_{t}= 11 mFg

_{c}= 13,600 NFg

_{t}= 69,700 N2. Homework Equations

μ = Ff/Fn

W

_{f}= F_{f}ΔDW

_{f}= μF_{g}ΔD = ΔK = 1/2mv^{2}μF

_{g}ΔD = 1/2mv^{2}μgΔD = 1/2v

^{2}v = sqrt(2μgΔD)

3. The Attempt at a Solution

(for car) v = sqrt(2μ

_{c}gΔD_{c})v

_{c}= sqrt(2(0.769)(9.81)(8.2)) = 11.1 m/s(for truck) v = sqrt(2μ

_{t}gΔD_{t})v

_{t}= sqrt(2(0.538)(9.81)(11)) = 10.8 m/sI 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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