91Ó°ÊÓ

• The stretch of the sling,$x=2.3m$.
• The escape velocity of the slingshot,$v=11.2/sor11200m/s$.
• Mass of the ball to be launched,$m=170gor0.17kg$.
• Force exerted by the average person, F =490N.

Using the formula of kinetic energy and potential energy, we can find the spring constant of the device if all the elastic potential energy is converted into kinetic energy. Using spring force we can find the total force and from this force and force exerted by a single person, we can find the number of people that are required to stretch the elastic bands.

Formulae:

The kinetic energy of the system,$KE=\frac{1}{2}M{V}^2$ (i)

The elastic potential energy of the system,$P{E}_{ELASTIC}=\frac{1}{2}K{X}^2$ (ii)

The force on a spring due to Hooke’s Law, $F=-KX$ (iii)

If elastic PE is converted into kinetic energy then we have the spring constant of the sling as:

$$\begin{gathered} P{E}_{elastic}=KE \\ \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2 \\ =k=\frac{m{v}^2}{x^2} \end{gathered}$$

role="math" localid="1657271549607" $$\begin{gathered} Asm=0.17kg,x=2.3m,v=11200m/s \\ k=\left(\frac{(0.17kg){(11200m/s)}^2}{{(2.3m)}^2}\right) \\ =4.03×{10}^{6}N/m \end{gathered}$$

Therefore, the spring constant of the device if all the elastic potential energy is converted into kinetic energy is$4.03×{10}^{6}N/M$ .

The total amount of spring force using equation (iii) is given as:

$$\begin{gathered} F=-\left(4.03×{10}^6\frac{N}{M}\right)×(2.3) \\ =-9.27×{10}^{6}N \end{gathered}$$

Considering the magnitude of the total force, as force exerted by an average person is $490$.

Therefore, number of people required to stretch the string:

$$\begin{gathered} n=\frac{Totalforce}{forceexertedbyeachperson} \\ =\frac{9.27×{10}^{6}N}{490N} \\ =18918.36persons \\ =18918persons \end{gathered}$$

Therefore, 18918 people are required to stretch the elastic bands.