Answer:
(D) F/2
Explanation:
Since the circular section is unbanked, the centripedal acceleration acting on each of the cars is
[tex]a_s = \frac{v_s^2}{r} = \frac{(2v)^2}{r} = \frac{4v^2}{r}[/tex]
[tex]a_l = \frac{v_l^2}{r} = \frac{v^2}{r}[/tex]
Therefore the centripetal force on each car
[tex]F_s = m_sa_s = \frac{4mv^2}{r}[/tex]
[tex]F_l = m_la_l = \frac{2mv^2}{r}[/tex]
Since [tex]F_s = 2F_l[/tex] this means the friction force required to keep the large car on the road is only half of the friction force required to keep the small car on road
So (D) F/2 is the correct answer
The frictional force required to keep the large car on the road without skidding is (D) F/2.
In this problem the centripetal force is provided by friction.
The centripetal force for any particle rotating in a circular section is given by equation (1)
[tex]F[/tex] = [tex]m\times V^2/r[/tex].....,,,(1)
where m = mass of particle
V = Velocity of the particle and
r = Radius of the Circle on which the particle is rotating
Given
There are two cars
Mass of small car = m
Speed of small car = 2v
so the frictional force for small car [tex]F_s[/tex] = [tex]m\times (2v)^2/r[/tex]...........(2)
Mass of large car = 2m
Speed of large car = v
so the frictional force for large car [tex]F_l[/tex] = [tex]2m\times (v)^2/r[/tex]...........(3)
On Dividing equation (3) by (2) we get
[tex]F_l[/tex] /[tex]F_s[/tex] = 1/2
[tex]F_l[/tex] = [tex]F_s[/tex] /2 = F/2 ( Given [tex]F_s[/tex] = F )
so
We can say that
Option D is correct.
For more information please refer the link below :
https://brainly.com/question/1623673
