1) Available force of friction: 6174 N
2) No
Explanation:
1)
The magnitude of the frictional force between the car's tires and the pavement of the road is given by
![F_f=\mu mg](https://tex.z-dn.net/?f=F_f%3D%5Cmu%20mg)
where
is the coefficient of friction
m is the mass of the car
g is the acceleration of gravity
For the car in this problem, we have:
(coefficient of friction)
m = 1260 kg (mass of the car)
![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
Therefore, the force of friction is
![F_f=(0.500)(1260)(9.8)=6174 N](https://tex.z-dn.net/?f=F_f%3D%280.500%29%281260%29%289.8%29%3D6174%20N)
2)
In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.
The centripetal force is given by
![F=m\frac{v^2}{r}](https://tex.z-dn.net/?f=F%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
m is the mass of the car
v is the tangential speed
r is the radius of the curve
In this problem, we have
m = 1260 kg
is the tangential speed
r = 41.6 m is the radius of the curve
Therefore, the centripetal force is
![F=(1260)\frac{15.0^2}{41.6}=6814 N](https://tex.z-dn.net/?f=F%3D%281260%29%5Cfrac%7B15.0%5E2%7D%7B41.6%7D%3D6814%20N)
Therefore, the force of friction is not enough to keep the car in the curve, since ![F_f](https://tex.z-dn.net/?f=F_f%3CF)