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A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.82, what is the greatest speed the car can have in the corner without skidding?

Respuesta :

anniwuanyanwu1

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

Lanuel

The greatest speed the car can have in the corner without skidding is 19.02 m/s.

Given the following data:

  • Mass of car = 1200 kg
  • Radius, r = 45 meters.
  • Coefficient of static friction = 0.82.

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]

To determine the greatest speed the car can have in the corner without skidding:

The greatest speed that a car will attain without skidding is given by the force of static friction and centripetal force:

Mathematically, the force of static friction is given by the formula;

[tex]F_s = uF_n[/tex]   ...equation 1.

Where;

  • Fs represents the force of static friction.
  • μ represents the coefficient of friction.
  • Fn represents the normal force.

Note: [tex]F_n = mg[/tex]

For centripetal force:

[tex]F_c = \frac{mv^2}{r}[/tex]   ....equation 2.

Equating the two (2) equations, we have:

[tex]umg = \frac{mv^2}{r}\\\\umgr=mv^2\\\\ugr = v^2\\\\v= \sqrt{ugr}[/tex]

Substituting the given parameters into the formula, we have;

[tex]v = \sqrt{0.82 \times 9.8 \times 45} \\\\v = \sqrt{361.62}[/tex]

Speed, v = 19.02 m/s

Read more: https://brainly.com/question/13754413

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