Question

A (hypothetical) large slingshot is stretched 1.53 m to launch a 130–g projectile with speed sufficient to escape from the Earth (11.2 km/s). (a) What is the force constant of the device, if all the potential energy is converted to kinetic energy? (b) Assume that an average person can exert a force of 220 N. How many people are required to stretch the slingshot?

Answer 1

Approximately
`6.97* 10^(6)\; {\rm N\cdot m^(-1)}`.

Approximately
`48,500`.

Step-by-step explanation:

The elastic potential energy (
`\text{EPE}`) stored in an ideal spring is:

`\displaystyle (\text{EPE}) = (1)/(2)\, k\, x^(2)`,

Where:

• 
  `k` is the spring constant (force constant), and
• 
  `x` is the distance by which the spring was compressed or stretched from the equilibrium position.

When an object of mass
`m` travels at a speed of
`v`, the kinetic energy (
`\text{KE}`) of that object would be:

`\displaystyle (\text{KE}) = (1)/(2)\, m\, v^(2)`.

In this question, it is assumed that the
`\text{EPE}` in the spring was entirely converted into the
`\text{KE}` of the projectile. Therefore:

`\displaystyle (\text{KE}) = (\text{EPE})`.

`\displaystyle (1)/(2)\, k\, x^(2) = (1)/(2)\, m\, v^(2)`.

Rearrange this equation to find the spring constant
`k`:

`\displaystyle k = (m\, v^(2))/(x^(2))`.

Apply unit conversion and ensure that each quantity is in standard units:

`\begin{aligned}v &= 11.2\; {\rm km\cdot s^(-1)} \\ &= 11.2 \; {\rm km\cdot s^(-1)} * \frac{10^(3)\; {\rm m}}{1\; {\rm km}} \\ &= 1.12 * 10^(4)\; {\rm m\cdot s^(-1)}\end{aligned}`.

`\begin{aligned}m &= 130\; {\rm g} = 0.130\; {\rm kg}\end{aligned}`.

It is given that the spring was stretched to
`x = 1.53\; {\rm m}`. Substitute the value of
`v`,
`m`, and
`x` into the expression for
`k` to obtain:

`\begin{aligned} k &= (m\, v^(2))/(x^(2)) \\ &= ((0.130)\, (1.12 * 10^(4))^(2))/((1.53)^(2))\; {\rm N\cdot m^(-1)} \\ &\approx 6.97 * 10^(6)\; {\rm N\cdot m^(-1)}\end{aligned}`.

To find the force required to stretch the spring, multiply the spring constant by the displacement from the equilibrium position:

`\begin{aligned} & k\, x \\ \approx\; & (6.9662 * 10^(6))\, (1.53)\; {\rm N} \\ \approx \; &1.0658 * 10^(7)\; {\rm N}\end{aligned}`.

Divide by
`220\; {\rm N}` to obtain:

`\displaystyle \frac{1.0658 * 10^(7)\; {\rm N}}{220\; {\rm N}} \approx 4.85 * 10^(4)`.

(Rounded up.)

In other words, it would require approximately
`48,500` people to stretch this spring.