91Ó°ÊÓ

Mass of car is m = 0.8kg

Radius of track is r = 5.00m

Normal force on car at point B is $N_B=6N$

The force applied to an item in curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force.

A fictitious force that moves in a circle and is directed away from the centre of the circle is called centrifugal force.

So the centrifugal force use the same formula of centripetal force but in negative sign

$F_c=-\frac{m{v}^2}{r}$

Where, m is mass of body, v is velocity of body and r is radius of circular path

A car is moving on a circular path so that a centrifugal force act in outward direction to the car.

Equating the forces at force at point B

$F_c=N_B+mg$ …(¾±)

Now take the point A and equating the forces

$N_A=mg+F_c$

From equation (i)

$$\begin{gathered} N_A=mg+N_B+mg \\ {N}_A=2mg+N_B \end{gathered}$$

Substitute all the values in above equation

$$\begin{gathered} N_A=\left(2×0.8kg×9.8m/s\right)+6N \\ =21.68N \end{gathered}$$

Hence the normal force on the car when it is at the bottom of the track (point A) is 21.68N