(II) A baseball (\(m = 145\;{\rm{g}}\)) traveling 32 m/s moves a fielder’s glove backward 25 cm when the ball is caught. What was the average force exerted by the ball on the glove?

The mass of the baseball is \(m = 145\;{\rm{g}} = 0.145\;{\rm{kg}}\).

The initial speed of the baseball is\({v_{\rm{i}}} = 32\;{\rm{m/s}}\).

The final speed of the baseball is\({v_{\rm{f}}} = 0\;{\rm{m/s}}\).

The fielder’s glove moves backward by a distance of \(d = 25\;{\rm{cm}} = 0.25\;{\rm{m}}\).

The initial kinetic energy of the baseball is \(\frac{1}{2}mv_{\rm{i}}^2\) .

The final kinetic energy of the baseball is zero as it comes to rest.

Therefore, the magnitude of the change in kinetic energy of the baseball is:

\(\begin{array}{c}\Delta K = \left| {0 - \frac{1}{2}mv_{\rm{i}}^2} \right|\\ = \frac{1}{2}mv_{\rm{i}}^2\end{array}\)

The magnitude of work done by the force exerted by the ball on the glove is equal to the magnitude of the change in kinetic energy of the ball.

Let F be the force exerted by the baseball on the glove.

The work done by the force is \(W = Fd\).

From the work-energy theorem, you can write:

\(\begin{array}{c}W = \Delta K\\Fd = \frac{1}{2}mv_{\rm{i}}^2\\F = \frac{{mv_{\rm{i}}^2}}{{2d}}\end{array}\)

Now, substituting the values in the above equation, you will get:

\(\begin{array}{c}F = \frac{{\left( {0.145\;{\rm{kg}}} \right) \times {{\left( {32\;{\rm{m/s}}} \right)}^2}}}{{2 \times 0.25\;{\rm{m}}}}\\ = 296.96\;{\rm{N}}\end{array}\)

Hence, the average force exerted by the baseball on the glove is 296.96 N.