91Ó°ÊÓ

The given data is listed below as:

• The velocity of the baseball is,v =46 m/s
• The mass of the baseball is,$\mathrm{m}=145\mathrm{g}×\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right)=145×{10}^{-3}\mathrm{kg}$
• The distance the pitcher exerted force is, s=1.0 m

This law states that the force applied by an object is directly proportional to the product of the acceleration and the mass of the object. The acceleration occurs as a force acts on a particular mass.

The equation of the acceleration of the ball is expressed as:

$$\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}$$

Here, v is the final speed, u is the initial speed, a is the acceleration of the ball, and s is the distance traveled by the ball.

As the ball was at rest initially, then the initial velocity is zero.

The equation of the force is expressed as:

$$\begin{gathered} F=ma \\ =m\frac{v^2-u^2}{2s} \end{gathered}$$

Here, F is the net force and m is the mass of the ball.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{F}=\left(145×{10}^{-3}\mathrm{kg}\right)\frac{{\left(46\mathrm{m}/\mathrm{s}\right)}^2-{\left(0\right)}^2}{2\left(1.0\mathrm{m}\right)} \\ =\left(145×{10}^{-3}\mathrm{kg}\right)\left(\frac{2116{\mathrm{m}}^2/{\mathrm{s}}^2}{\left(2\mathrm{m}\right)}\right) \\ =\left(145×{10}^{-3}\mathrm{kg}\right)\left(1058\mathrm{m}/{\mathrm{s}}^2\right) \\ =153\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, further as:

$$\begin{gathered} \mathrm{F}=153\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2×\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =153\mathrm{N} \end{gathered}$$

Thus, the force produced on the ball is 153 N.

The free-body diagram of the ball has been drawn below:

In the above diagram, the weight of the ball is acting downwards. The normal forcen is directed upwards. The forceF is directed in the right direction, and the accelerationa is directed in the right direction.