Question

A swimmer moves through the water at an average speed of 0.22 m/s. The average drag force is 110 N. What average power is required of the swimmer?

Short answer

The stopping distance of an automobile is 100 m.

Step-by-step solution

Given data:

Weight of the passengers, W = 16400 N

Average speed of an automobile, v = 113 km/h

The frictional on the wheels from the road, $f_k=8230\mathrm{N}$

To understand the concept:

Using the formula for workdone on a system by external force you can find thestopping distance of an automobile.

Formula:

$W=∆E_{mech}+∆E_{th}$

Calculate the initial speed of the automobile:

Mass of the passengers is,

$$\begin{gathered} m=\frac{W}{g} \\ =\frac{16400}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =1673\mathrm{kg} \end{gathered}$$

Average speed of an automobile is,

$$\begin{gathered} v=113km/h \\ =113\times \left(\frac{5}{18}\right)\mathrm{m}/\mathrm{s} \\ =31.4\mathrm{m}/\mathrm{s} \end{gathered}$$

Calculate the average power required:

Work done on a banana-Earth system by external force is given by,

$W=∆E_{mech}+∆E_{th}$

In this case , W= 0

Then,

$$\begin{gathered} 0=∆E_{mech}+∆E_{th} \\ 0=∆KE+∆PE+∆E_{th} \\ 0=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+0+f_{k}d \\ {f}_{k}d=\frac{1}{2}\left(v_i^2-v_f^2\right) \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} d\left(8230\mathrm{N}\right)=\frac{1}{2}\left(1673\mathrm{kg}\right){\left(31.4\right)}^2 \\ d=100\mathrm{m} \end{gathered}$$

Hence, the stopping distance of an automobile is 100 m .