Chapter 11.1, Problem 1PP

To determine

The skater’s initial kinetic energy, the amount of energy that is transformed into other forms of energy as she stops, and the work that must be done by her to speed up again to $2.5\text{m/s}$ .

Answer to Problem 1PP

  $162.5\text{}\text{J}$ , $162.5\text{}\text{J}$ , and $162.5\text{}\text{J}$ , respectively

Explanation of Solution

Given:

The mass of the skater, $m=52.0\text{kg}$

Initial velocity of the skater, $v_i=2.5\text{}\text{m/s}$

The distance covered after she glides to stop, $d=24.0\text{m}$

Formula used:

The kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $KE=\frac{1}{2}m{v}^2$

And, the work done on an object is given by the change in its kinetic energy,

  $W=K{E}_f-K{E}_i$

Where $K{E}_f$ is the final kinetic energy, and $K{E}_i$ is the initial kinetic energy.

Calculation:

Using the given values, the initial kinetic energy of the skater will be,

  $\begin{array}{c}K{E}_i=\frac{1}{2}m{v}^2\\ =\frac{1}{2}\left(52\right){\left(2.5\right)}^2\\ =162.5\text{}\text{J}\end{array}$

Now, the amount of energy that is transformed into other forms of energy as she stops will be equal to the change in her kinetic energy as she stops, thus

  $\begin{array}{c}{E}_{\text{transformed}}=K{E}_f-K{E}_i\\ =\left(\frac{1}{2}m{\left(0\right)}^2-162.5\right)\text{J}\\ =-162.5\text{}\text{J}\end{array}$

Negative sign denotes that the skater’s energy is reduced.

Similarly, the work that must be done by her to speed up again to $2.5\text{m/s}$ will be,

  $\begin{array}{c}W=\left(KE\right){\text{'}}_f-\left(KE\right){\text{'}}_i\\ =\left(\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(0\right)}^2\right)\\ =\frac{1}{2}\left(52\right){\left(2.5\right)}^2-0\\ =162.5\text{}\text{J}\end{array}$

Conclusion:

Thus, the skater’s initial kinetic energy is $162.5\text{}\text{J}$ , the amount of energy that is transformed into other forms of energy as she stops is $162.5\text{}\text{J}$ , and the work that must be done by her to speed up again to $2.5\text{m/s}$ is $162.5\text{}\text{J}$ .