A 13 500 N car traveling at 50.0 km/h rounds a curve of radius 2.00 × 102 m. Find the following: a. the centripetal acceleration of the car b. the centripetal force c. the minimum coefficient of static friction between the tires and the road that will allow the car to round the curve safely

Respuesta :

Answer:

a. 0.947 m/s^2

b. 1304.54 N

c. 0.0966

Explanation:

mass of car = 13500 N = 13500/9.8 = 1377.55 kg

velocity = 50 km/h = 50,000 m/h = 13.9 m/s

raidus = 204 m

a. centripetal acceleartion = v^2/r = 13.9^2/204 = 0.947 m/s^2

b. centripetal force = m*centripetal acceleration = 1377.55 * 0.947 = 1304.54 N

c. In order for the car to round the curve safely, static friction = centripetal force

static friction = coefficient of friction (mu) * mg = mu* 1377.55*9.8 = 13500mu

13500mu = 1304.54

mu = 1304.54/13500 = 0.0966

The acceleration, force and coefficient of friction is required.

Centripetal acceleration is [tex]0.965\ \text{m/s}^2[/tex]

Centripetal force is [tex]1328\ \text{N}[/tex]

Coefficient of friction is [tex]0.1[/tex]

N = Weight of car = 13500 N

v = Velocity = [tex]50=\dfrac{50}{3.6}=13.89\ \text{m/s}[/tex]

r = Radius = [tex]2\times 10^2\ \text{m}[/tex]

m = Mass of car = [tex]\dfrac{N}{g}[/tex]

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Centripetal acceleration is

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{13.89^2}{2\times 10^2}\\\Rightarrow a_c=0.965\ \text{m/s}^2[/tex]

Force is given by

[tex]F_c=ma_c\\\Rightarrow F_c=\dfrac{N}{g}a_c\\\Rightarrow F_c=\dfrac{13500}{9.81}\times 0.965\\\Rightarrow F_c=1328\ \text{N}[/tex]

Coefficient of friction is given by

[tex]\mu=\dfrac{F_c}{N}\\\Rightarrow \mu=\dfrac{1328}{13500}\\\Rightarrow \mu=0.098\approx 0.1[/tex]

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