pezmachine9000
Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)
If the coefficient of static friction between the tires of the car and the surface of the road is μ[tex]_{s}[/tex] = 0.402, then what is the maximum speed at which the car can safely exit the highway without sliding?

The Radius Of Curvature Of A Highway Exit Is R 935 M The Surface Of The Exit Road Is Horizontal Not Banked See Figure If The Coefficient Of Static Friction Betw class=

Sagot :

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r   ⇒   v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.