91Ó°ÊÓ

The given data can be listed below as,

• The mass of the car is, $m=1.60\text{ k²µ}$.
• The speed of the car is,$v=12\text{″¾/²õ}$ .
• The radius of the track is,$r=5.00\text{″¾}$ .

The normal force mainly a contact force, if the two surface are in contact than there is a normal force and if they are not in contact than there is no normal force.

Part (a)

The expression for the magnitude of normal force using centripetal force can be expressed as,

$\begin{array}{l}{F}_A=mg+\frac{m{v}^2}{r}\\ F_A=m(g+\frac{v^2}{r})\end{array}$

Here $m$is the mass of the car,$g$ is the acceleration due to gravity $v$, is the speed, and$r$ is the radius of the track.

Substitute $1.60\text{ k²µ}$for $m$,$9.8{\text{″¾/²õ}}^{\text{2}}$for $g$,$12\text{″¾/²õ}$for $v$, and 5.00 m for $r$in the above equation.

$\begin{array}{c}{F}_A=1.60\text{ k²µ}×(9.8{\text{″¾/²õ}}^{\text{2}}+\frac{{\left(12{\text{″¾/²õ}}^{\text{2}}\right)}^2}{5.00\text{″¾}})\\ =61.76\text{ N}\end{array}$

Hence, required magnitude of normal force at point A is, 61.76 N.

Part (b)

The expression for the magnitude of normal force using centripetal force can be expressed as,

$\begin{array}{l}{F}_B=mg−\frac{m{v}^2}{r}\\ F_B=m(g−\frac{v^2}{r})\end{array}$

Substitute$1.60\text{ k²µ}$ for $m$, $9.8{\text{″¾/²õ}}^{\text{2}}$for $g$,$12\text{″¾/²õ}$ for$v$ , and 5.00 m for$r$ in the above equation.

$\begin{array}{c}{F}_B=1.60\text{ k²µ}×(9.8{\text{″¾/²õ}}^{\text{2}}−\frac{{\left(12{\text{″¾/²õ}}^{\text{2}}\right)}^2}{5.00\text{″¾}})\\ =−30.4\text{ N}\end{array}$

Hence, required magnitude of normal force at point B is, -30.4 N.