Drag race tires in contact with an asphalt surface probably have one of the highest coefficients of static friction in the everyday world. Given a car with a mass of 3000kg:

Assuming a constant acceleration and no slipping of tires, what acceleration does the car experience if it covers a quarter mile in 6.0 s? (1 mi = 1609 m)

What is the coefficient of static friction? (Assume the magnitude of the force that accelerates the car is approximately equal to the magnitude of the frictional force between the tires and the road.)

Did you need the mass to calculate the coefficient of friction? Why or why not?

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CS9537 CS9537 · Answer 1 · 2017-12-03T04:55:56+01:00

From s = ½at²
a = 2s / t² = 2 * ¼ * 1609m / (6.0s)² = 22 m/s² ◄ acceleration

µ = a / g = 22 / 9.8 = 2.3

both answers to two significant digits.

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aristocles aristocles · Answer 2 · 2019-08-24T05:08:05+02:00

Answer:

Part a)

[tex]a = 22.35 m/s^2[/tex]

Part b)

[tex]\mu = 2.28 [/tex]

Part c)

No, mass is required to solve for the friction coefficient

Explanation:

As we know that car is moving at constant acceleration

so here we can say that the distance traveled is given as

[tex]d = \frac{1}{2}at^2[/tex]

here we know

[tex]d = \frac{1609}{4} = 402.25 m[/tex]

t = 6 s

now we have

[tex]402.25 = \frac{1}{2}(a)(6^2)[/tex]

[tex]a = 22.35 m/s^2[/tex]

Part b)

now we know that this acceleration is due to frictional force

so we have

[tex]F_f = \mu mg[/tex]

[tex]ma = \mu mg[/tex]

[tex]a = \mu g[/tex]

[tex]22.35 = \mu (9.81)[/tex]

[tex]\mu = 2.28 [/tex]

Part c)

No, mass is required to solve for the friction coefficient