91Ó°ÊÓ

The weight of a baseball is $m=9.0\mathrm{oz}$.

The initial speed of baseball $v_1=81.8\mathrm{mi}/\mathrm{h}$.

The final speed of baseball $v_2=110\mathrm{ft}/\mathrm{s}$.

Use the equation of change in mechanical energy making sure that all quantities are in the FPS (British) system of units.

Formula is as follow:

$∆E_{mec}=\left(K_1+U_1\right)-\left(K_2+U_2\right)$

where, $∆E_{mec}$is mechanical energy, K is kinetic energy and U is potential energy.

Converttheunits of all quantities in FPS system,

$$\begin{gathered} weight=mg \\ =9.0\mathrm{oz} \\ =9.0\mathrm{oz}×\frac{1\mathrm{lb}}{16\mathrm{oz}} \\ =0.5\mathrm{lb} \end{gathered}$$

So, the mass of the baseball is,

$$\begin{gathered} m=\frac{0.56\mathrm{lb}}{32\mathrm{t}/{\mathrm{s}}^2} \\ =0.0175\mathrm{lb}·{\mathrm{s}}^2/\mathrm{ft} \\ =0.0175\mathrm{slug} \end{gathered}$$

Also, the initial velocity of a baseball is 81.8 mi/h, so,

$$\begin{gathered} v_1=81.8\frac{\mathrm{mi}}{\mathrm{h}}=\frac{81.8\mathrm{mi}}{1\mathrm{hour}}×\frac{5280\mathrm{ft}}{1\mathrm{mi}}×\frac{1\mathrm{hour}}{3600\mathrm{s}} \\ =119.97\mathrm{ft}/\mathrm{s} \end{gathered}$$

Now, the equation for change in mechanical energy as,

$∆E_{mec}=\left(K_1+U_1\right)-\left(K_2+U_2\right)$

As the height of ball isthesame for initial and final position, there is no change in the potential energy. Hence, the above equation becomes,

$$\begin{gathered} ∆E_{mec}=K_1-K_2 \\ =\frac{1}{2}m\left(v_1^2-v_2^2\right) \\ ∆E_{mec}=\frac{1}{2}×0.0175×\left(119.{97}^2-{110}^2\right) \\ ∆E_{mec}=20.06\mathrm{ft}·\mathrm{lb} \\ =20\mathrm{ft}·\mathrm{lb} \end{gathered}$$

Hence,the amount of decrease in mechanical energy of theball-earth system is$∆E_{mec}=20\mathrm{ft}·\mathrm{lb}$.

Therefore, the decrease in mechanical energy can be found using the initial and final values of velocity.