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A 1305 kg test car travels around a flat circular track that is 2.15 km long (circumference). If the coefficient of friction between the car’s tires and the track is 1.20 will the car be able to drive a speed of 55.7 m/s and stay on the track?

User VHow
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Final answer:

Yes, the car will be able to drive at a speed of 55.7 m/s and stay on the track.

Step-by-step explanation:

To determine whether the car will be able to drive at a certain speed and stay on the track, we need to calculate the centripetal force exerted on the car. The formula for centripetal force is Fc = m*v^2/r, where Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the circular track.

Using the given values, we have:

Mass (m) = 1305 kg

Speed (v) = 55.7 m/s

Radius (r) = circumference / 2π = 2150 m

Plugging these values into the formula, we get:

Fc = (1305 kg) * (55.7 m/s)^2 / 2150 m = 19061.4 N

The centripetal force (Fc) must be provided by the frictional force between the tires and the track. The maximum frictional force is given by the equation Ffric = μ * Fn, where μ is the coefficient of friction and Fn is the normal force.

The normal force (Fn) is equal to the weight of the car, which is given by the equation Fn = m * g, where g is the acceleration due to gravity.

Using the given values:

Fn = (1305 kg) * (9.8 m/s^2) = 12759 N

Now we can calculate the maximum frictional force:

Ffric = (1.20) * (12759 N) = 15310.8 N

Since the calculated centripetal force (19061.4 N) is greater than the maximum frictional force (15310.8 N), the car will be able to drive at a speed of 55.7 m/s and stay on the track.

User CCondron
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