A highway curves to the left with radius of curvature of 36 m and is banked at 28 ◦ so that cars can take this curve at higher speeds. Consider a car of mass 1717 kg whose tires have a static friction coefficient 0.87 against the pavement. top view R = 36 m 28 ◦ rear view µ = 0.87 How fast can the car take this curve without skidding to the outside of the curve? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

30.34m/s

Explanation:

The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg

Sum of forces in the x direction

mv²/r = N sin 28 + μN cos 28

mv²/r = N(sin 28 + μcos 28)

Thus,

mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]

v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]

v²/36 = 9.8 (1.2376/0.4744)

v²/36 = 9.8 * 2.6088

v²/36 = 25.57

v² = 920.52

v = 30.34m/s

it is the second option.

the rearrangement of particles in a physical change. hope this