Final answer:
The greatest speed a car with a mass of 1.00×103 kg can achieve without slipping on a 50.0 m radius track with a coefficient of friction of 0.600 is approximately 17.15 meters per second.
Step-by-step explanation:
To find the greatest speed a 1.00×103 kg car can attain without slipping on a level unbanked circular track with a coefficient of friction of 0.600 and a radius of 50.0 m, we must first understand that the maximum frictional force that can act on the car is what provides the centripetal force necessary to keep the car moving in a circle. This force can be calculated using the equation for the force of friction: fmax = μN, where μ is the coefficient of friction, and N is the normal force. In this scenario, the normal force (N) is equal to the gravitational force on the car, which is the mass (m) times the acceleration due to gravity (g), or N = mg.
The centripetal force (Fc) required to keep the car in circular motion can be described by the equation Fc = mv2/r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the curve. Since the frictional force is what provides the centripetal force, we can set these two equations equal to each other and solve for v:
μN = mv2/r → μmg = mv2/r → v2 = μrg → v = √(μrg)
By plugging in the given values for μ (0.600), r (50.0 m), and g (9.80 m/s2), we can find the velocity:
v = √(0.600 × 50.0 m × 9.80 m/s2)
v = √(294 m2/s2)
v ≈ 17.15 m/s
Therefore, the greatest speed at which the car can travel without slipping is approximately 17.15 meters per second.