(II) What is the maximum speed with which a 1200 -kg car can round a turn of radius 90.0 m on a flat road if the coefficient of friction between tires and road is 0.65 ? Is this result independ ent of the mass of the car?

## Step 1<br />The problem involves the concept of circular motion and friction. The car is moving in a circular path, and the friction between the tires and the road provides the necessary force to keep the car moving in this path. The maximum speed at which the car can take the turn without skidding is determined by the force of friction.<br /><br />## Step 2<br />The force of friction is given by the formula:<br />### \(F_{\text{friction}} = \mu \cdot F_{\text{normal}}\)<br />where \(\mu\) is the coefficient of friction and \(F_{\text{normal}}\) is the normal force, which is equal to the weight of the car.<br /><br />## Step 3<br />The force required to keep the car moving in a circular path is given by the formula:<br />### \(F_{\text{centripetal}} = \frac{m \cdot v^2}{r}\)<br />where \(m\) is the mass of the car, \(v\) is the velocity of the car, and \(r\) is the radius of the circular path.<br /><br />## Step 4<br />At the maximum speed, the force of friction equals the centripetal force. Therefore, we can set the two equations equal to each other and solve for \(v\):<br />### \(\mu \cdot m \cdot g = \frac{m \cdot v^2}{r}\)<br /><br />## Step 5<br />Solving this equation for \(v\), we get:<br />### \(v = \sqrt{\mu \cdot g \cdot r}\)<br /><br />## Step 6<br />The mass of the car cancels out in the equation, which means that the maximum speed at which the car can take the turn is independent of the mass of the car.