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A 1300 kg car drives around a flat 170-m -diameter circular track at 22 m/s .

What minimum coefficient of static friction is needed to keep the car from sliding off the road?

User KarlM
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1 Answer

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Final answer:

To keep the car from sliding off the road, a minimum coefficient of static friction of 0.54 is needed.

Step-by-step explanation:

To find the minimum coefficient of static friction needed to keep the car from sliding off the road, we can use the equation:

Fc = m × v2 / r

Where:

- Fc is the centripetal force

- m is the mass of the car

- v is the velocity of the car

- r is the radius of the circular track

Plugging in the values given, we have:

Fc = (1300 kg) × (22 m/s)2 / (85 m)

= 6,724 N

Next, we can calculate the maximum static friction force using the equation:

fs = μs × N

Where:

- fs is the static friction force

- μs is the coefficient of static friction

- N is the normal force

The normal force can be found as:

N = m × g

Where g is the acceleration due to gravity.

The maximum static friction force would occur when the car is on the verge of sliding off the road, meaning fs = Fc. Therefore:

Fc = μs × N

Substituting the known values, we get:

6,724 N = μs × (1300 kg × 9.8 m/s2)

Solving for μs, we find:

μs = 0.54

Therefore, a minimum coefficient of static friction of 0.54 is needed to keep the car from sliding off the road.

User Roman Meyer
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