Chapter 3, Problem 22A

The three pucks are sliding across ice at the noted speeds. Air resistance and ice friction are negligible.

a. Rank them, from greatest to least, by the force needed to keep them going.

b. Rank them, from greatest to least, by the force needed to stop them in the same time interval.

To determine

Ranking of pucks from greatest to least, by the force needed to keep them going.

Answer to Problem 22A

Force acting on the three pucks is zero: $F_A=F_B=F_C=0$ .

Explanation of Solution

Given:

Air resistance and ice friction are negligible.

Speed of puck-A is $v_A=2\text{ m/s}$

Speed of puck-B is $v_B=4\text{ m/s}$

Speed of puck-C is $v_C=6\text{ m/s}$

Description:

According to the first law of equilibrium: ${\displaystyle \sum F}=0$

As given in the statement, air resistance and ice friction are negligible. So, there will be no horizontal force acting on the puck. Hence, for all the three pucks, there is no external force acting. So, the force acting on each puck is zero.

  $F_A=F_B=F_C=0$

Conclusion:

Force acting on the three pucks is zero: $F_A=F_B=F_C=0$

To determine

Ranking of pucks from greatest to least, by the force needed to stop them.

Answer to Problem 22A

Ranking of pucks from greatest to least, by the force needed to stop them is $F_C>F_B>F_A$ .

Explanation of Solution

Given:

Air resistance and ice friction is negligible.

Speed of puck-A is $u_A=2\text{ m/s}$

Speed of puck-B is $u_B=4\text{ m/s}$

Speed of puck-C is $u_C=6\text{ m/s}$

Formula used:

Newton’s second law is given by: $F=ma$

Calculation:

According to the statement, the final velocity will be zero. Now, using the above formula, we get:

  $\begin{array}{l}F=m\left(\frac{v-u}{t}\right)\\ F=m\left(\frac{0-u}{t}\right)\\ F\propto u\end{array}$

So, the force exerted is directly proportional to the initial speed, given that the mass and time interval is constant. Hence, the ranking is given by:

  $u_C>u_B>u_A$

  $F_C>F_B>F_A$

Conclusion:

Ranking of pucks from greatest to least, by the force needed to stop them is $F_C>F_B>F_A$ .