Chapter 13.1, Problem 13.12P

A package is thrown down an incline at A with a velocity of 1 m/s. The package slides along the surface ABC to a conveyor belt that moves with a velocity of 2 m/s. Knowing that $d=6$m and $\mu k=0.2$ between the package and all surfaces, determine (a) the speed of the package at C. (b) the distance the package will slide on the conveyor belt before it comes to rest relative to the belt.

To determine

(a)

To calculate:

The speed of package atC.

Answer to Problem 13.12P

The speed of package at C: $v_c=3.464m/s$

Explanation of Solution

Given information:

Velocity atA: $v_A=1m\text{/}s$

Friction coefficient between the package and the surfaceABC: ${\mu }_k=0.2$

Velocity of belt: $v_{belt}=2m\text{/}s$

Distance betweenAandB, $d=6m$

Distance betweenBandC, $x_{BC}=7m$

Calculations:

Draw the free body diagram of the block on inclineAB:

$\begin{array}{l}{N}_{AB}=mg\cos {30}^{\circ }\\ F_{AB}={\mu }_k{N}_{AB}=0.2mg\cos {30}^{\circ }\\ U_{A\to B}=mgd\sin {30}^{\circ }-F_{AB}d\\ U_{A\to B}=mgd\left(\sin {30}^{\circ }-{\mu }_k\cos {30}^{\circ }\right)\end{array}$

Free body diagram of the block on level surfaceBC:

$\begin{array}{l}{N}_{BC}=mg\\ F_{BC}={\mu }_{k}mg\\ U_{B\to C}=-{\mu }_{k}mg{x}_{BC}\\ \Rightarrow AtA,\\ T_A=\frac{1}{2}m{v}_A^{2}and{v}_A=1m\text{/}s\\ \Rightarrow AtC,\\ T_C=\frac{1}{2}m{v}_C^{2}and{v}_C=2m\text{/}s\end{array}$

Assume that there is no loss of energy at cornerB.

Applying concept of Work and energy:

$T_A+U_{A\to B}+U_{B\to C}=T_C$

$\frac{1}{2}m{v}_A^2+mgd\left(\sin {30}^{\circ }-{\mu }_k\cos {30}^{\circ }\right)-{\mu }_{k}mg{x}_{BC}=\frac{1}{2}m{v}_0^2$

Solving the equation for $v_C^2$ ,

$\begin{array}{l}{v}_C^2=v_A^2+2gd\left(\sin {30}^{\circ }-{\mu }_k\cos {30}^{\circ }\right)-2{\mu }_{k}g{x}_{BC}\\ v_C^2={\left(1\right)}^2+\left(2\right)\left(9.81\right)\left(6\right)\left(\sin {30}^{\circ }-0.2\cos {30}^{\circ }\right)-\left(2\right)\left(0.2\right)\left(9.81\right)\left(7\right)\\ v_C^2=12.002m^2/s^2\\ \text{Hence,}\\ v_c=3.464m/s\end{array}$

Conclusion:

The answer is calculated by a trigonometric equation and concept of work and energy.

To determine

(b)

To calculate:

The distance a package will slide on a conveyor belt before it comes to rest relative to the belt.

Answer to Problem 13.12P

The distance travelled by the package: $x_{belt}=2.039m$

Explanation of Solution

Given information:

Friction coefficient between the packages and the surfaceABC: ${\mu }_k=0.2$

Velocity of belt: $v_{belt}=2m\text{/}s$

Speed of package atC: $v_c=3.464m/s$

Calculations:

For the motion of box on the conveyor belt:

Let $x_{belt}$ is the distance travelled by a package when it slides on the conveyor belt.

$\begin{array}{l}+\uparrow \sum F_y=m{a}_y,\\ N-mg=0\\ \text{or, }N=mg\\ F_x={\mu }_{k}N={\mu }_{k}mg\end{array}$

Applying the concept of work and energy:

$\begin{array}{l}\frac{1}{2}m{v}_C^2-{\mu }_{k}mg{x}_{belt}=\frac{1}{2}m{v}_{belt}^2\\ \text{Solving for }{x}_{belt},\end{array}$

$\begin{array}{l}{x}_{belt}=\frac{v_C^2-v_{belt}^2}{2{\mu }_{k}g}\\ x_{belt}=\frac{12.002-2^2}{\left(2\right)\left(0.2\right)\left(9.81\right)}\\ x_{belt}=2.039m\end{array}$

Conclusion:

The distance travelled by the package is calculated by the concept of work and energy and equating the values of mv2, mg and mv2 belt.