Chapter 5.2, Problem 26PP

To determine

The distance traveled by disk before it comes to rest.

Whether the shot will reach 10-point section of the board or not.

Answer to Problem 26PP

The distance traveled by disk before it comes to rest is 6.95 m.

The disk will not reach the 10-point section.

Explanation of Solution

Given:

The initial velocity of disk, $u=6.5\text{m/s}$

The distance between the releasing point & 10-point section, $d=7.2\text{m}$

Formula Used:

From the Newton’s second law:

  $\begin{array}{l}{\displaystyle \sum F}=-ma\\ F_f=-ma\\ {\mu }_{k}mg=-ma\\ a=-{\mu }_{k}g\end{array}$

Here, $F_f$ represents force of friction,

  $m$ represents the mass of the disk,

  $a$ represents the acceleration due to gravity of the disk,

  ${\mu }_k$ represents kinetic coefficient of friction between disk and surface.

From the Newton’s equation of motion:

  $\begin{array}{l}{v}^2-u^2=2as\\ s=d_f-d_i=d_f-0\\ d_f=\frac{v^2-u^2}{2a}\end{array}$

Here, $v$ represents final velocity of the disk,

  $s$ represents the distance traveled,

  $a$ represents the acceleration due to gravity of the disk,

  $u$ represents the initial velocity of the disk,

  $d_f$ represents the final distance to which disk moved,

  $d_i$ represents the initial distance from which disk started.

Calculation:

Calculate the acceleration using the Newton’s second law:

  $\begin{array}{c}a=-{\mu }_{k}g\\ =-0.31(9.81)\\ =-3.04{\text{m/s}}^2\end{array}$

Calculate the distance using the Newton equation of motion:

  $\begin{array}{c}{d}_f=\frac{0^2-{6.5}^2}{2(-3.04)}\\ =\frac{42.25}{6.08}\\ =6.95\text{m}\end{array}$

As the value of distance travel is less than the distance to reach at 10-point section on the board.

So, it will not reach the 10-point on the board.

Conclusion:

The distance traveled by disk before it comes to rest is 6.95 m. Thus, the disk will not reach the 10-point section.