Q37P - University Physics with Modern Physics

Question

The fastest pitched baseball was measured at 46m/s. A typical baseball has a mass of 145 g. If the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, (a) what force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcher’s hand.

Step-by-Step Solution

Answer

(a) The force produced on the ball is 153 N.

(b) .

Step 1: Identification of the given data

The given data is listed below as:

 

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

Step 2: Significance of the second law of Newton

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.

Step 3: (a) Determination of the force produced on the ball

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.

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

Hence, further as:

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

 

 

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

Step 4: (b) Determination of the free body diagram of the ball

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 force n is directed upwards. The force F is directed in the right direction, and the acceleration a is directed in the right direction.