Step 1

We are given the mass of package $m = {\rm{15}}\;{\rm{kg}}$, the coefficient of static friction as ${\mu _s} = 0.8$, and the initial velocity of the conveyor belt as $u = 4\;{\rm{m/s}}$.

We are asked to determine shortest time the belt can stop.

 
Step 2

We will draw a free body diagram of the package.

Here, $f$ is the friction force, and $N$ is the normal force.

 
Step 3

We will find the friction force between the belt and the package.

\[f = {\mu _s}N\]

Substitute the given value in the above equation.

\[f = 0.8N\;\;\;...\left( 1 \right)\]
 
Step 4

We will resolve the forces in the perpendicular direction of the belt.

\[\begin{array}{c} N - mg\cos 30^\circ = 0\\ N = mg\cos 30^\circ \end{array}\]

Here, $g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ is the gravitational acceleration.

Substitute the given value in the above equation.

\[\begin{array}{c} N = \left( {{\rm{15}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\cos 30^\circ \\ = \left( {{\rm{15}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\cos 30^\circ \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ \approx 127.4\;{\rm{N}} \end{array}\]
 
Step 5

We will resolve the forces along the belt.

\[\begin{array}{c} f - mg\sin 30^\circ = - ma\\ 0.8N - mg\sin 30^\circ = - ma\\ 0.8N = mg\sin 30^\circ - ma \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} 0.8N = \left( {{\rm{15}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin 30^\circ - \left( {{\rm{15}}\;{\rm{kg}}} \right)a\\ 0.8\left( {127.4\;{\rm{N}}} \right) \times \left( {\frac{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{{\rm{1}}\;{\rm{N}}}}} \right) = \left( {{\rm{15}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin 30^\circ - \left( {{\rm{15}}\;{\rm{kg}}} \right)a\\ a = - \frac{{0.8\left( {127.4\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) - \left( {{\rm{15}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin 30^\circ }}{{{\rm{15}}\;{\rm{kg}}}}\\ \approx - 1.9\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 6

We will find the shortest time the belt can stop the belt.

\[v = {u_0} + at\]

Here, $v = 0\;{\rm{m/s}}$ is the final velocity of the package.

Substitute the given value in the above equation.

\[\begin{array}{c} 0\;{\rm{m/s}} = 4\;{\rm{m/s}} + \left( { - 1.9\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)t\\ t \approx 2.1\;{\rm{s}} \end{array}\]