Chapter 13.1, Problem 13.12P

Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to conveyor belt which moves with a velocity of 2 m/s. Knowing that $d=7.5$ m and ${\mu }_k=0.25$ between the packages and all surfaces, determine (a) the speed of the package at C, (b) the distance a package will slide on the conveyor belt before it comes to rest relative to the belt.

To determine

(a)

The speed of package at point $C$.

Answer to Problem 13.12P

The speed of package at point $C$ is $2.9\text{ m/s}$.

Explanation of Solution

Given:

Velocity of packages at point $A$ is $1\text{ m/s}$.

Velocity of conveyor is $2\text{ m/s}$.

Coefficient of kinetic friction is $0.25$.

Angle of inclination of conveyor is $30°$.

Concept used:

Refer to Fig P13.12; consider the movement of package from point $A$ to point $B$.

Draw FBD of block at point A.

Write the expression for the kinetic energy of the package at point $A$.

$T_A=\frac{1}{2}m{v}_A^2$   ...... (1)

Here, $T_A$ is the kinetic energy of the package at point $A$, $m$ is the mass of the package and $v_A$ is the initial velocity of the package.

Write the expression for the kinetic energy of the package at point $B$.

$T_B=\frac{1}{2}m{v}_B^2$   ...... (2)

Here, $T_B$ is the kinetic energy of the package at point $B$ and $v_B$ is the velocity of the package at point $B$.

Write the expression for the work done for moving package from point $A$ to point $B$.

$U_{AB}=\left(mg\sin \theta -F\right)\left(d\right)$   ...... (3)

Here, $U_{AB}$ is the work done to move package from point $A$ to point $B$, $g$ is the acceleration due to gravity, $\theta$ is the angle of inclination, $d$ is the distance moved by the package from $A$ to $B$ and $F$ is the frictional force acting on the package.

For the system to be in static equilibrium the summation of all the forces in the direction of motion should be zero.

$\sum F=0$

Take summation of forces in the direction of motion.

$\sum F=N-mg\cos \theta$

Substitute $0$ for $\sum F$ in the above expression and rearrange for $N$.

$N=mg\cos \theta$

Write the expression for the force of friction acting on the package.

$F={\mu }_{k}N$

Substitute $mg\cos \theta$ for $N$ in the above expression.

$F={\mu }_k\left(mg\cos \theta \right)$

Substitute ${\mu }_k\left(mg\cos \theta \right)$ for $F$ in the equation (3).

$U_{AB}=\left(mg\sin \theta -{\mu }_k\left(mg\cos \theta \right)\right)\left(d\right)$   ...... (4)

Write the expression as per work energy principle for package.

$T_A+U_{AB}=T_B$   ...... (5)

Draw the FBD of the package at point $B$ ,

Write the expression for the kinetic energy at point $C$.

$T_C=\frac{1}{2}m{v}_C{}^2$   ...... (6)

Here, $T_C$ is the kinetic energy of package at point $C$ and $v_C$ is the velocity of package at point $C$.

Write the expression for the work done for moving package from point $B$ to point $C$.

$U_{BC}=-{\mu }_{k}mgd\text{'}$   ...... (7)

Here, $U_{BC}$ is the work done for moving the package from $B$ to $C$, $d\text{'}$ is the distance between the points $B$ and $C$.

Write the expression as per work energy principle for package.

$T_B+U_{BC}=T_C$   ...... (8)

Calculation:

Substitute $1\text{ m/s}$ for $v_A$ is equation (1).

$\begin{array}{c}{T}_A=\frac{1}{2}m{\left(1\right)}^2\\ =0.5m\end{array}$

Substitute $9.81{\text{ m/s}}^2$ for $g$, $30°$ for $\theta$, $7.5\text{ m}$ for $d$ and $0.25$ for ${\mu }_k$ in equation (4).

$\begin{array}{c}{U}_{AB}=\left(m\left(9.81\right)\sin 30°-\left(\mathrm{0.252.781}\right)\left(m\left(9.81\right)\cos 30°\right)\right)\left(7.5\right)\\ =20.858m\end{array}$

Substitute $\frac{1}{2}m{v}_B{}^2$ for $T_B$, $0.5m$ for $T_A$ and $20.858m$ for $U_{AB}$ in equation (5).

$\begin{array}{l}0.5m+20.858m=\frac{1}{2}m{v}_B{}^2\\ \left(0.5+20.858\right)m=\frac{1}{2}m{v}_B{}^2\end{array}$

Rearrange the above expression for $v_B$.

$\begin{array}{c}{v}_B=\sqrt{42.716}\\ =6.535\text{ m/s}\end{array}$

Substitute $0.25$ for ${\mu }_k$, $9.81{\text{ m/s}}^2$ for $g$ and $7\text{ m}$ for $d\text{'}$ in equation (7).

$\begin{array}{c}{U}_{BC}=-\left(0.25\right)m\left(9.81\right)\left(7\right)\\ =-17.168m\end{array}$

Substitute $\frac{1}{2}m{v}_B{}^2$ for $T_B$, $\frac{1}{2}m{v}_C{}^2$ for $T_C$ and $-17.168m$ for $U_{BC}$ in equation (8).

$\begin{array}{l}\frac{1}{2}m{v}_B{}^2-17.168m=\frac{1}{2}m{v}_C{}^2\\ 0.5v_B{}^2-17.168=0.5v_C{}^2\end{array}$

Substitute $6.535\text{ m/s}$ for $v_B$ in above expression.

$\begin{array}{l}0.5{\left(6.535\right)}^2-17.168=0.5v_C{}^2\\ 21.353-17.168=0.5v_C{}^2\end{array}$

Simplify the above expression for $v_C$.

$\begin{array}{c}{v}_C=\sqrt{8.37}\\ =2.893\text{ m/s}\\ \approx \text{2}\text{.9}\text{m/s}\end{array}$

The speed of package at point $C$ is $2.9\text{ m/s}$.

Conclusion:

Thus, the speed of package at point $C$ is $2.9\text{ m/s}$.