b) The maximum speed of the car is 47.0 m/s
a) The centripetal force on the car is 7363 N
Explanation:
b)
We have to proceed by solving part b) first. The frictional force between the tires of the car and the road provides the centripetal force that keeps the car in circular motion, therefore we can write:
[tex]\mu mg = m\frac{v^2}{r}[/tex]
where the term on the left is the frictional force while the term on the right is the centripetal force, and where
[tex]\mu = 0.75[/tex] is the coefficient of friction
m = 1000 kg is the mass of the car
r = 300 m is the radius of the turn
v is the speed of the car
The maximum speed of the car is the speed for which the frictional force is still enough to keep the car in circular motion, so
[tex]\mu mg \geq m\frac{v^2}{r}[/tex]
Therefore, solving for [tex]v[/tex], we find
[tex]v = \sqrt{\mu g r}=\sqrt{(0.75)(9.8)(300)}=47.0 m/s[/tex]
a)
The centripetal force for an object in circular motion is given by
[tex]F=m\frac{v^2}{r}[/tex]
where
m is the mass of the object
v is its speed
r is the radius of the circular path
For the car in this problem,
m = 1000 kg
v = 47.0 m/s
r = 300 m
Solving for F, we find the force:
[tex]F=(1000) \frac{47.0^2}{300}=7363 N[/tex]
Learn more about circular motion:
brainly.com/question/2562955
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