During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke’s law with a spring constant of $100\,N/m$. If the hose is stretched by $5.00\,m$ and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

Hint:The catapult obeys Hooke's law since it uses the mechanism of elasticity to throw an object over a distance. The purpose of the catapult is to launch an object into air which travels through projectile motion. The concept of Hooke’s law is taken into consideration since a force is used to throw the balloon from the catapult. The formula for work done by this force is applied to determine the work done.

Formula used:
The work done by a body exhibiting elastic properties is given by:
$W = \dfrac{1}{2} \times k \times {\left( {{x_i} - {x_f}} \right)^2}$
Where, $W$ is the work done, $k$ is the spring constant, ${x_i}$ is extended length and ${x_f}$ is the original length.

Complete step by step answer:
A catapult is a device used to launch an object through the air. This is done by pulling its hose along with the object held together backwards and then releasing it, sending it flying across the air. The house of the catapult is made up of rubber material which has a property known as elasticity. Elasticity is the property by which the body regains its original size and shape after a force that deforms it, has been applied on it. Since it is elastic it means that the hose can be easily stretched from its original position.

This principle is what is used in catapults wherein there is an external force applied in pulling the hose backwards, that is, in the direction opposite of the direction you want the object (which in this case is the water balloon) to be launched in and hence it can be seen that there is some displacement of the hose from its original position.

This is where Hooke’s law comes into picture since there is a force applied on an elastic material along with a displacement. Hooke's law states that the force that is applied in order to extend or compress a body is directly proportional to the amount of displacement of the body. This is given by the following equation:
$F = - kx$
Now, let us extract the data given in the question. The constant of proportionality, that is, the spring constant and the length till which the hose is stretched, that is, its extension is also given.

Given, $k = 100\;N/m$
$ \Rightarrow x = 5m$
Since there is a force that is applied to launch the water balloon by stretching or displacing the hose from its original position there is some amount of work done in this process. This amount of work done is what we have been asked to find. Work is said to be done when there is a force applied on a body and the body moves through some distance due to the force exerted on it.

This work is stored in the form of some energy and this is the energy that is used to launch the balloon up in the air. The work done on this elastic body is stored in the form of potential energy. This energy is the reason due to which the balloon is launched in air. The work done or potential energy stored by the house of the catapult is given by an equation:
$W = \dfrac{1}{2} \times k \times {\left( {{x_i} - {x_f}} \right)^2}$

When the hose undergoes an extension there is a displacement from its relaxed position which is indicated by ${x_f}$ to its extended position indicated by variable ${x_i}$.
Here the initial position is assumed to be zero because there is no displacement that occurs when the hose is in its relaxed position. Hence the equation becomes:
$W = \dfrac{1}{2} \times k \times {\left( {{x_i} - 0} \right)^2}$
$ \Rightarrow W = \dfrac{1}{2} \times k \times x_i^2$ --------($1$)
Now the above given values are substituted in equation ($1$) in order to get the work done by the house of the catapult. By substituting we get:
$W = \dfrac{1}{2} \times 100 \times {5^2}$
$ \Rightarrow W = \dfrac{1}{2} \times 100 \times 25$
$ \Rightarrow W = \dfrac{{2500}}{2}$
$ \therefore W = 1250\;J$

Therefore the amount of work done on the hose in the pouch to launch the water balloon in air before it reaches its relaxed length is $1250\;J$.

Note:The negative sign in the equation for Hooke’s law does not indicate its magnitude since the sign only denotes the direction of pull of the elastic body which opposes the direction the object should move in. The hose of a catapult is always pulled in the direction opposite to the direction it is required to be thrown in.