a) [tex]0.68 m/s^2[/tex]
b) 964.5 N
c) 0.069
Explanation:
a)
When an object is moving in a circular motion, the direction of its velocity is changing - therefore, it has an acceleration towards the center of the circle, called centripetal acceleration.
The magnitude of the centripetal acceleration is given by
[tex]a=\frac{v^2}{r}[/tex]
where
v is the speed of the object
r is the radius of the circle
For the car in this problem:
v = 40.0 km/h = 11.1 m/s is the speed
r = 180 m is the radius of the circle
Substituting, we find the acceleration:
[tex]a=\frac{11.1^2}{180}=0.68 m/s^2[/tex]
b)
The centripetal force is the force that keeps the object along its circular motion. It also acts towards the center of the circle, and it is given by
[tex]F=ma[/tex]
where
m is the mass of the object
a is the centripetal acceleration
Here the weight of the car is
[tex]W=mg=13,900 N[/tex]
where
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
So the mass is
[tex]m=\frac{W}{g}=\frac{13,900}{9.8}=1418.4 kg[/tex]
Therefore, the centripetal force is
[tex]F=(1418.4)(0.68)=964.5 N[/tex]
c)
In this case, the force of static friction between the tires and the road provides the required centripetal force to keep the car in circular motion. This force is given by:
[tex]F_f=\mu mg[/tex]
where
[tex]\mu[/tex] is the coefficient of friction
Equating the frictional force to the centripetal force,
[tex]\mu mg=ma[/tex]
So we get:
[tex]\mu=\frac{a}{g}[/tex]
And substitutng:
[tex]a=0.68 m/s^2[/tex] (centripetal acceleration)
[tex]g=9.8 m/s^2[/tex]
We find:
[tex]\mu=\frac{0.68}{9.8}=0.069[/tex]
