Chapter 5, Problem 81QAP

To determine

(a)

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.100.

Answer to Problem 81QAP

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.100 is 8.12 s.

Explanation of Solution

Given:

The length of the ski slope

  $\Delta x=250\text{ m}$

The angle made by the slope to the horizontal

  $\theta =37.0°$

Initial speed of the ski

  $v_0=10.0\text{ m/s}$

Coefficient of kinetic friction between the ski and snow

  ${\mu }_k=0.100$

Formula used:

A free body diagram of the ski 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 ski 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 to obtain the value of the time t.

  $\Delta x=v_{0}t+\frac{1}{2}{a}_x{t}^2$......(5)

Calculation:

Substitute the values of the variables in equation (4) and calculate the value of the acceleration.

  $\begin{array}{c}{a}_x=g\left(\sin \theta -{\mu }_k\cos \theta \right)\\ =\left(9.80{\text{ m/s}}^2\right)\left[\left(\sin 37.0°\right)-\left(0.100\right)\left(\cos 37.0°\right)\right]\\ =5.12{\text{ m/s}}^2\end{array}$

Use the calculated value of a_x and the values of v₀ and $\Delta x$ in equation (5).

  $\left(250\text{ m}\right)=\left(10.0\text{ m/s}\right)t+\frac{1}{2}\left(5.12{\text{ m/s}}^2\right)t^2$

Solve the quadratic equation.

  $\left(2.56{\text{ m/s}}^2\right)t^2+\left(10.0\text{ m/s}\right)t+\left(-250\text{ m}\right)=0$

  $\begin{array}{c}t=\frac{-\left(10.0\text{ m/s}\right)\pm \sqrt{{\left(10.0\text{ m/s}\right)}^2-4\left(2.56{\text{ m/s}}^2\right)\left(-250\text{ m}\right)}}{2\left(2.56{\text{ m/s}}^2\right)}\\ =\frac{-\left(10.0\text{ m/s}\right)\pm \left(51.58\text{ m/s}\right)}{\left(5.12{\text{ m/s}}^2\right)}\end{array}$

Taking the positive root,

  $t=8.12\text{ s}$

Conclusion:

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.100 is 8.12 s.

To determine

(b)

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.150.

Answer to Problem 81QAP

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.150 is 8.39 s.

Explanation of Solution

Given:

The length of the ski slope

  $\Delta x=250\text{ m}$

The angle made by the slope to the horizontal

  $\theta =37.0°$

Initial speed of the ski

  $v_0=10.0\text{ m/s}$

Coefficient of kinetic friction between the ski and snow

  ${\mu }_k=0.150$

Formula used:

The acceleration of the ski down the slope is given by

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

The time taken to reach the bottom of the slope is calculated using the expression,

  $\Delta x=v_{0}t+\frac{1}{2}{a}_x{t}^2$

Calculation:

Substitute the values of the variables in equation for acceleration and calculate the value of the acceleration.

  $\begin{array}{c}{a}_x=g\left(\sin \theta -{\mu }_k\cos \theta \right)\\ =\left(9.80{\text{ m/s}}^2\right)\left[\left(\sin 37.0°\right)-\left(0.150\right)\left(\cos 37.0°\right)\right]\\ =4.72{\text{ m/s}}^2\end{array}$

Use the calculated value of a_x and the values of v₀ and $\Delta x$ in equation of motion.

  $\left(250\text{ m}\right)=\left(10.0\text{ m/s}\right)t+\frac{1}{2}\left(4.72{\text{ m/s}}^2\right)t^2$

Solve the quadratic equation.

  $\left(2.36{\text{ m/s}}^2\right)t^2+\left(10.0\text{ m/s}\right)t+\left(-250\text{ m}\right)=0$

  $\begin{array}{c}t=\frac{-\left(10.0\text{ m/s}\right)\pm \sqrt{{\left(10.0\text{ m/s}\right)}^2-4\left(2.36{\text{ m/s}}^2\right)\left(-250\text{ m}\right)}}{2\left(2.36{\text{ m/s}}^2\right)}\\ =\frac{-\left(10.0\text{ m/s}\right)\pm \left(49.6\text{ m/s}\right)}{\left(4.72{\text{ m/s}}^2\right)}\end{array}$

Taking the positive root,

  $t=8.39\text{ s}$

Conclusion:

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.150 is 8.39 s.