Question

(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?

Short answer

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

Step-by-step solution

Given data

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 \(d = 25\;{\rm{cm}} = 0.25\;{\rm{m}}\).

Calculation of the change in kinetic energy

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:

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

Calculation of the average force

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.