Chapter 6, Problem 27PQ

Curling is a game similar to lawn bowling except it is played on ice and instead of rolling balls on the lawn, stones are slid along ice. A curler slides a stone across a sheet of ice with an initial speed v_i in the positive x direction. The coefficient of kinetic friction between the stone and the curling lane is μ_k. Express your answers in terms of v_i, μ_k, and g only. a. What is the acceleration of the stone as it slides down the lane? b. What distance does the curling stone travel?

To determine

The acceleration of the stone as it slides down the lane.

Answer to Problem 27PQ

The acceleration of the stone as it slides down the lane is $-{\mu }_{k}g\widehat{\text{i}}$

Explanation of Solution

Take horizontal direction to be $x$ direction and vertical direction to be $y$ direction.

The stone slides on a horizontal surface so that the net force in the vertical direction must be zero.

Write the expression for the Newton’s second law of motion in $y$ direction.

  $\Sigma F_y=0$                                                                                                                      (I)

Here, $\Sigma F_y$ is the magnitude of net force in $y$ direction.

Write the expression for $\Sigma F_y$ .

  $\Sigma F_y=F_N-mg$

Here, $F_N$ is the magnitude of the normal force, $m$ is the mass of the stone and $g$ is the acceleration due to gravity.

Put the above equation in equation (I).

  $\begin{array}{c}{F}_N-mg=0\\ F_N=mg\end{array}$                                                                                                          (II)

Write the expression for the Newton’s second law of motion in $x$ direction.

  $\Sigma F_x=m{a}_x$

Here, $\Sigma F_x$ is the magnitude of net force in $x$ direction and $a_x$ is the magnitude of the acceleration of the stone.

Rewrite the above equation for $a_x$ .

  $a_x=\frac{\Sigma F_x}{m}$                                                                                                                 (III)

After the stone loses contact with the curling stick, the only force acting in the horizontal direction is the force of friction. The force of friction acts opposite to the motion of the stone.

Write the expression for $\Sigma F_x$ .

  $\Sigma F_x=-F_f$                                                                                                               (IV)

Here, $F_f$ is the magnitude of the force of friction.

Write the expression for $F_f$ .

  $F_f={\mu }_k{F}_N$

Here, ${\mu }_k$ is the coefficient of kinetic friction.

Put equation (II) in the above equation.

  $F_f={\mu }_{k}mg$

Put the above equation in equation (IV).

  $\Sigma F_x=-{\mu }_{k}mg$

Put the above equation in equation (III).

  $\begin{array}{c}{a}_x=\frac{-{\mu }_{k}mg}{m}\\ =-{\mu }_{k}g\end{array}$                                                                                                             (V)

The acceleration acts along $x$ direction. Write the expression for the acceleration.

  ${\overrightarrow{\text{a}}}_x=a_x\widehat{\text{i}}$                                                                                                                   (VI)

Here, ${\overrightarrow{\text{a}}}_x$ is the acceleration of the stone.

Conclusion:

Put equation (V) in equation (VI).

  ${\overrightarrow{\text{a}}}_x=-{\mu }_k\text{g}\widehat{\text{i}}$

Therefore, the acceleration of the stone as it slides down the lane is $-{\mu }_{k}g\widehat{\text{i}}$

To determine

The distance travelled by the curling stone.

Answer to Problem 27PQ

The distance travelled by the curling stone is $\Delta x=\frac{v_i^2}{2{\mu }_{k}g}$.

Explanation of Solution

Write the Newton’s constant acceleration equation.

  $v_f^2=v_i^2+2a_x\Delta x$

Here, $v_f$ is the final velocity of the stone, $v_i$ is the initial velocity of the stone and $\Delta x$ is the distance travelled by the stone.

Rewrite the above equation for $\Delta x$ .

  $\Delta x=\frac{v_f^2-v_i^2}{2a_x}$

Put equation (V) in the above equation.

  $\begin{array}{c}\Delta x=-\frac{v_f^2-v_i^2}{2{\mu }_{k}g}\\ =\frac{v_i^2-v_f^2}{2{\mu }_{k}g}\end{array}$                                                                                                       (VII)

Conclusion:

The final velocity of the stone is zero.

Substitute $0$ for $v_f$ in equation (VII) to find $\Delta x$ .

  $\begin{array}{c}\Delta x=\frac{v_i^2-0}{2{\mu }_{k}g}\\ =\frac{v_i^2}{2{\mu }_{k}g}\end{array}$

Therefore, the distance travelled by the curling stone is $\Delta x=\frac{v_i^2}{2{\mu }_{k}g}$.