Q28E - University Physics with Modern Physics

Question

A soccer ball with mass 0.420 kg is initially moving with speed 2 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball’s motion. Over what distance must the player’s foot be in contact with the ball to increase the ball’s speed to  6.00 m/s

Step-by-Step Solution

Answer

The distance travelled by the player to increase the velocity is 0.168 m.

Step 1: Identification of given data

The given data can be listed below,

• The mass of the soccer ball is, m = 0.420 kg.
• The initial speed of soccer is, $v_i=2\hspace{0.33em}m/s$.
• The constant force acting on the ball is, $F_c=40\hspace{0.33em}N$.
• The final speed of the ball is, $v_f=6.0\hspace{0.33em}m/s$. 

Step 2: Concept/Significance of work done.

According to the work-energy theorem, a system's change in energy is what constitutes work. A moving object's kinetic energy is altered when it is stopped. Therefore work has been done on it.

Step 3: Determination of the distance must the player’s foot be in contact with the ball to increase the ball’s speed

The total work done on the ball is given by,

 

$$\begin{gathered} W_{tot}=K_2-K_1 \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \end{gathered}$$                                                                                                    …(i)

 

The work done by the force is given by,

$$\begin{gathered} W_{tot}=\overrightarrow{f}·\overrightarrow{s} \\ =fs \cos \varphi \end{gathered}$$                                                                                                                  …(ii)

 

Compare equation (i) and (ii), the distance travelled by the player is given by,

$$\begin{gathered} fs \cos \varphi =\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2 \\ s=\frac{1}{2f}m\left(v_f^2{-}_i^2\right) \end{gathered}$$ 

 

Substitute all the values in the above,

 

$$\begin{gathered} s=\frac{1}{2f}m\left(v_f^2{-}_i^2\right) \\ =\frac{1}{2\left(40 N\right)}\left(0.42 kg\right)\left[{\left(6 m/s\right)}^2-{\left(2 m/s\right)}^2\right] \\ =\frac{13.44}{40}m \\ =0.168 m \end{gathered}$$ 

 

Thus, the distance traveled by the player to increase the velocity is 0.168 m.