Question

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?

Short answer

The work by the hose on the balloon is $W_{\text{s}}=1250\text{J}$

Step-by-step solution

Given data

1. The spring constant of the hose is,$k=100\text{N/m}$ .
2. The maximum elongation of the hose is,$x_i=5.00\text{m}$.
3. The final position of the hose is, $x_f=0\text{m}$.

Understanding the concept

We can use the concept of work done by the spring on the block.

Formulae:

$W_{\text{s}}=\frac{1}{2}k{x}_{\text{i}}^{\text{2}}-\frac{1}{2}k{x}_{\text{f}}^{\text{2}}$

Calculate how much work the force from the hose does on the balloon

Work energy theorem states that change in the kinetic energy of the system will be equal to the work done of the system.

The work by the spring is,

$W_{\text{s}}=\frac{1}{2}k\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)$

Substitute the values in the above expression, and we get,

$$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times \left(100\text{N/m}\right)\times \left[{\left(5.00\text{m}\right)}^2-{\left(0\text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times \left(100\right)\times {\left(5.00\right)}^2\left(1\text{N/m}\times 1{\text{m}}^{\text{2}}\times \frac{1\text{J}}{1\text{N}·\text{m}}\right) \\ {W}_{\text{s}}=1250\text{J} \end{gathered}$$

Thus, the work by the hose on the balloon is $W_{\text{s}}=1250\text{J}$.