1. The frictional force
When an object moves in a circular path, we said that it is moving by circular motion.
In order to do so, the object must be acted upon a force that keeps the object changing its direction and "turning" along the circular path. This force is called centripetal force, and its direction is always towards the centre of the circular path.
The nature of this centripetal force is always different, and depends on the situation. For example, for a satellite orbiting the Earth, the centripetal force is provided by the gravitational attraction between the Earth and the satellite.
For a car moving round a curve, such as in this problem, the centripetal force is provided by the force of friction between the tyres of the car and the road, which acts towards the centre of the turn and keeps the car moving in circular motion.
2. 2000 N
The magnitude of the centripetal force is given by
[tex]F=m\frac{v^2}{r}[/tex]
where
m is the mass of the car
v is its speed
r is the radius of the curve
For the car in this problem,
m = 500 kg
r = 100 m
v = 20 m/s
Substituting, we find the force:
[tex]F=(500)\frac{20^2}{100}=2000 N[/tex]
3. 18000 N
As before, the magnitude of the centripetal force is given by
[tex]F=m\frac{v^2}{r}[/tex]
This time however, the speed of the car is
v = 60 m/s
While the other data are unchanged
m = 500 kg
r = 100 m
So, the new centripetal force on the car is
[tex]F=(500)\frac{60^2}{100}=18000 N[/tex]
4. The force has increased by a factor 9
We can understand this part by looking again at the equation:
[tex]F=m\frac{v^2}{r}[/tex]
We notice that the centripetal force is proportional to the square of the speed:
[tex]F\propto v^2[/tex]
In this problem, the speed of the car has been tripled from part 2) to part 3):
[tex]v_1 = 20 m/s\\v_2 = 60 m/s = 3 v_1[/tex]
Therefore, the new force will increase by a factor [tex]3^2 =9[/tex]:
[tex]F_2 \propto v_2 ^2 = (3v_1)^2 = 9 v_1^2 \rightarrow F_2 = 9 F_1[/tex]
