Chapter 5, Problem 82QAP

To determine

The coefficient of kinetic friction between the package and the inclined plane, so that the package reaches the bottom with no speed.

Answer to Problem 82QAP

The coefficient of kinetic friction between the package and the inclined plane, so that the package reaches the bottom with no speed is 0.382.

Explanation of Solution

Given info:

The mass of the package

  $m=2.50\text{-kg}$

Length of the inclined plane

  $\Delta x=12.0\text{ m}$

Angle made by the plane to the horizontal

  $\theta =20.0°$

Initial speed of the package

  $v_0=2.00\text{ m/s}$

Final speed of the package

  $v=0\text{ m/s}$

Formula used:

A free body diagram of the package is drawn to analyze its motion.

  

Assume a coordinate system, with the +x directed downwards along the incline and +y directed upwards, perpendicular to the incline. The weight $\overrightarrow{w}$ acts vertically downwards, the normal force $\overrightarrow{n}$ acts perpendicular to the incline along the +y direction. The force of kinetic friction ${\overrightarrow{f}}_k$ acts upwards along the incline along the −x direction.

Resolve the weight $\overrightarrow{w}$ along the +x and the −y directions. Use the expression $w=mg$, where g is the acceleration of free fall and write expressions for the components.

  $\begin{array}{l}{w}_x=w\sin \theta =mg\sin \theta \\ w_y=w\cos \theta =mg\cos \theta \end{array}$...... (1)

The package is in equilibrium along the y direction.

Therefore,

  ${\displaystyle \sum F_y}=n-w_y=0$

Therefore, using equation (1),

  $n=w_y=mg\cos \theta$.....(2)

The force of kinetic friction and the normal force are related according to the following equation:

  $f_k={\mu }_{k}n$

From equation (2)

  $f_k={\mu }_{k}n={\mu }_{k}mg\cos \theta$.....(3)

Write the force equation along the +x direction.

  ${\displaystyle \sum F_x}=w_x-f_k=m{a}_x$

Use the values of w_x and f_k from equations (2) and (3) in the expression,

  $\begin{array}{c}{w}_x-f_k=m{a}_x\\ mg\sin \theta -{\mu }_{k}mg\cos \theta =m{a}_x\end{array}$

Simplify and write an expression for a_x.

  $a_x=g\left(\sin \theta -{\mu }_k\cos \theta \right)$......(4)

Use the following equation of motion with equation (4) to determine the coefficient of kinetic friction.

  $v^2=v_0^2+2a_x\Delta x$.....(5)

Calculation:

Use the values of the variables in equation (5) and determine the package s acceleration.

  $\begin{array}{c}{v}^2=v_0^2+2a_x\Delta x\\ {\left(0\text{ m/s}\right)}^2={\left(2.00\text{ m/s}\right)}^2+2a_x\left(12.0\text{ m}\right)\\ a_x=\frac{{\left(0\text{ m/s}\right)}^2-{\left(2.00\text{ m/s}\right)}^2}{2\left(12.0\text{ m}\right)}\\ =-1.67{\text{ m/s}}^2\end{array}$

Rewrite equation (4) for ${\mu }_k$.

  ${\mu }_k=\frac{g\sin \theta -a_x}{g\cos \theta }$

Substitute the value of a_x in the above equation along with the values of other variables and calculate the value of ${\mu }_k$.

  $\begin{array}{c}{\mu }_k=\frac{g\sin \theta -a_x}{g\cos \theta }\\ =\frac{\left(9.80{\text{ m/s}}^2\right)\left(\sin 20.0°\right)-\left(-1.67{\text{ m/s}}^2\right)}{\left(9.80{\text{ m/s}}^2\right)\left(\cos 20.0°\right)}\\ =0.382\end{array}$

Conclusion:

The coefficient of kinetic friction between the package and the inclined plane, so that the package reaches the bottom with no speed is 0.382.