Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:
[tex] a_{c} = \frac{v^{2}}{r} [/tex]
Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m
[tex] a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2} [/tex]
Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:
[tex] F_{c} = ma_{c} [/tex]
Where:
m: is the mass of the car
The mass is given by:
[tex] P = m*g [/tex]
Where P is the weight of the car = 13561 N
[tex] m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg [/tex]
Now, the centripetal force is:
[tex] F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N [/tex]
Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:
[tex] F_{c} = F_{\mu} [/tex]
[tex] F_{c} = \mu N = \mu P [/tex]
[tex] \mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069 [/tex]
Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!
