# 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.

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}