A (hypothetical) large slingshot is stretched 2.30 m to launched a 170 g projectile with speed sufficient to escape from Earth `(11.2 km//s)`. Assume the elastic bands of the slingshot obey Hooke's law.
Assume that an average person can be exert a force of 490 N. How many people are required to stretch the elastic bands?

Options

A. `9.5 xx 10^(3)`
B. `1.89 xx 10^(4)`
C. `3.78 xx 10^(4)`
D. `2.85 xx 10^(4)`

Answer

To solve the problem of how many people are required to stretch the elastic bands of a slingshot, we will follow these steps:

### Step 1: Understand the problem
We have a slingshot that stretches 2.30 m to launch a projectile of mass 170 g (0.17 kg) at a speed of 11.2 km/s (which is the escape velocity). We need to find out how many people are needed to exert the force required to stretch the elastic bands.

### Step 2: Convert units
Convert the mass of the projectile from grams to kilograms and the speed from km/s to m/s.
- Mass of the projectile, \( m = 170 \, \text{g} = 0.17 \, \text{kg} \)
- Speed of the projectile, \( v = 11.2 \, \text{km/s} = 11.2 \times 10^3 \, \text{m/s} \)

### Step 3: Use energy conservation
The elastic potential energy stored in the slingshot when stretched is equal to the kinetic energy of the projectile when launched.

The elastic potential energy (EPE) is given by:
\[ EPE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the stretch (2.30 m).

The kinetic energy (KE) of the projectile is given by:
\[ KE = \frac{1}{2} m v^2 \]

### Step 4: Set the energies equal
Setting the elastic potential energy equal to the kinetic energy:
\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]
We can cancel the \( \frac{1}{2} \) on both sides:
\[ k x^2 = m v^2 \]

### Step 5: Solve for the spring constant \( k \)
Rearranging the equation gives:
\[ k = \frac{m v^2}{x^2} \]
Substituting the values:
\[ k = \frac{0.17 \, \text{kg} \times (11.2 \times 10^3 \, \text{m/s})^2}{(2.30 \, \text{m})^2} \]

Calculating \( v^2 \):
\[ v^2 = (11.2 \times 10^3)^2 = 1.2544 \times 10^8 \, \text{m}^2/\text{s}^2 \]

Calculating \( x^2 \):
\[ x^2 = (2.30)^2 = 5.29 \, \text{m}^2 \]

Now substituting these values into the equation for \( k \):
\[ k = \frac{0.17 \times 1.2544 \times 10^8}{5.29} \approx 4.0 \times 10^6 \, \text{N/m} \]

### Step 6: Calculate the force required to stretch the slingshot
Using Hooke's law, the force \( F \) required to stretch the slingshot is given by:
\[ F = k x \]
Substituting the values:
\[ F = 4.0 \times 10^6 \, \text{N/m} \times 2.30 \, \text{m} = 9.2 \times 10^6 \, \text{N} \]

### Step 7: Determine the number of people required
If one person can exert a force of 490 N, the number of people \( N \) required is:
\[ N = \frac{F}{\text{Force per person}} = \frac{9.2 \times 10^6 \, \text{N}}{490 \, \text{N}} \approx 18878.57 \]
Rounding up, we find that approximately 18879 people are required to stretch the elastic bands.

### Final Answer
**Approximately 18879 people are required to stretch the elastic bands of the slingshot.**

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