Formulae: $v_x=d\frac{x}{d}t$:$,a_x=d\frac{v_x}{d}t$ or $a_x=v_{x}d\frac{v_x}{d}x$;

$v=d\frac{s_l}{d}t$;$a_n=\frac{v^2}{}$;$,a_t=d\frac{v}{d}t$;$,x_B-x_A+$xBA;

${\displaystyle \underset{i}{\overset{?}{}}}{F}_{}=m{a}_x$;$,F_{fk}={}_{k}N$;$,F_{fs}{}_{s}N$;$,F_{\text{e}\text{l}\text{x}}=-kx$

At the instant shown, the $1\mathrm{.}5-$kg packages ride on the surface of the conveyor belt.

The coefficient of friction $($static friction$)$ between the belt and a package is ${}_s=0\mathrm{.}45.$ The radius of curvature for the round part of the belt is $15\mathrm{.}2$

$cm=0\mathrm{.}152m=6in$

Neglect the size of the packages.

Draw the free body diagram of the package on the inclined surface $AB$ if the angle $$ is $20.(2$ points$)$

Find the magnitude of the normal reaction force acting on the package on the inclined surface $AB$ if the angle $$ is $20.(2$ points$)$

Find the value of the friction force acting on the package on the inclined surface $AB$ if the belt starts from rest with the constant acceleration and its speed increases to $1\frac{m}{s}$ in $2$ s $.(1$ point$)$

If the belt is moving at its constant speed of $1\frac{m}{s},$ draw the free body diagram of the package on the round surface of the belt at the point where angle $$ is $5.(2$ points$)$

Find the package's normal and tangential accelerations at the point where angle $$ is $5($on the round surface of the belt$)$ if the belt is moving at its constant speed of $1\frac{m}{s}.(1$ point$)$

Find the value of the friction force acting on the package at the point where angle $$ is $5($on the round surface of the belt$)$ if the belt is moving at its constant speed of $1\frac{m}{s}.(1$ point$)$

At what angle $$ do the packages first begin to slip off the surface of the belt after the belt is moving at its constant speed of $1\frac{m}{s}?(1$ point$)$