# A 17-kg suitcase falls from a hot-air balloon that is floating at a height of 333 m above the...

## Question:

A 17-kg suitcase falls from a hot-air balloon that is floating at a height of 333 m above the surface of Earth. The suitcase has an initial speed of 0 m/s but it reaches a speed of 26 m/s just before it hits the ground. Calculate the percentage of the initial energy that is "lost" to air resistance.

## Law of Conservation of Energy:

Let us consider a medium that consists of some amount of energy. Here, the given medium is considered as perfectly isolated. In this scenario, the amount of trapped energy will be conserved forever.

## Answer and Explanation:

**Given data:**

- Mass of the suitcase, {eq}m = 17 \ kg {/eq}

- Initial speed, {eq}u = 0 {/eq}

- Final speed, {eq}v = 26 \ m/s {/eq}

- Height, {eq}h = 333 \ m {/eq}

Let the magnitude of the energy lost due to air resistance be ** E**.

From the energy conservation law,

{eq}\begin{align*} E &= E_{top} - E_{bottom}\\ E &= (mgh + \frac{1}{2} mu^{2}) - (mgh_{0} + \frac{1}{2} mv^{2})\\ E &= (17 \times 9.80 \times 333 + \frac{1}{2} \times 17 \times (0)^{2}) - (17 \times 9.80 \times 0 + \frac{1}{2} \times 17 \times (26)^{2})\\ E &= 49731.8 \ J.\\ \end{align*} {/eq}

Here, energy at the top is given as,

{eq}\begin{align*} E_{top} &= mgh + \frac{1}{2} mu^{2}\\ E_{top} &= 17 \times 9.80 \times 333 + \frac{1}{2} \times 17 \times (0)^{2}\\ E_{top} &= 55477.8 \ \rm J.\\ \end{align*} {/eq}

Thus, percentage energy lost in this case can be given as,

{eq}\begin{align*} \left ( \frac{E}{E_{top}} \right ) \% &= \frac{49731.8}{55477.8} \times 100\\ \left ( \frac{E}{E_{top}} \right ) \% &= 89.64 \%\\ \end{align*} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6