A car has a mass of 2500 kg rounds a circular turn with a radius of 50 m. If the road is flat and the coefficient of friction is 0.7 between the tires and the road, how fast can the car go without skidding?
9m/s
To determine the maximum speed at which the car can go without skidding, we need to consider the forces acting on the car. The two main forces to consider are the force of gravity pulling the car downward and the frictional force between the tires and the road.
The force of gravity can be calculated using the equation Fg = m * g, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s²). In this case, Fg = 2500 kg * 9.8 m/s² = 24,500 N.
The frictional force can be calculated using the equation Ff = μ * N, where μ is the coefficient of friction and N is the normal force. In this case, N is equal to the force of gravity (N = Fg), and μ = 0.7. Therefore, Ff = 0.7 * Fg.
Now, the centripetal force required to keep an object moving in a circular path can be calculated using the equation Fc = m * v² / r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the circular turn.
In this case, Fc = m * v² / r.
Since the frictional force provides the centripetal force, we can equate Ff to Fc:
0.7 * Fg = m * v² / r
Rearranging the equation to solve for v:
v² = 0.7 * Fg * r / m
v = √(0.7 * Fg * r / m)
Plugging in the given values:
v = √(0.7 * 24,500 N * 50 m / 2500 kg)
v = √(8,575,000 Nm / 2500 kg)
v = √(3420 m²/s²)
v = 58.5 m/s
Therefore, the car can go without skidding at a maximum speed of approximately 58.5 m/s.