Question

A book prone to air resistance is released from rest 300 m

above the ground. It has a speed of 40 m/s when it reaches the
ground. Assume U = 0 on the ground, what percentage of the
initial mechanical energy of the book-Earth system was
dissipated due to air resistance?​

Answer 1

Approximately
`73\%`.

(Assuming that
`g = \rm 9.81\; m \cdot s^(-2)`.)

Step-by-step explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let
`m(\text{book})` represent the mass of the book.

The initial potential energy of the book is

`\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 * 300) \cdot m(\text{book})\cr &= \left(2.943* 10^3\right) \cdot m(\text{book})\end{aligned}`.

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be
`(2.943* 10^3) \cdot m(\text{book})`.

When the book was about to hit the ground, its speed is
`\rm 40\; m \cdot s^(-1)`. Its kinetic energy would be:

`\begin{aligned} \text{KE} &= (1)/(2) \, m(\text{book}) \cdot v^(2) \cr &= \left((1)/(2) * 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00* 10^2\right)\cdot m(\text{book})\end{aligned}`.

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be
`\left(8.00* 10^2\right)\cdot m(\text{book})` when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

`\begin{aligned}&\left(2.943* 10^3\right) \cdot m(\text{book}) - \left(8.00* 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143* 10^3\right)\cdot m(\text{book})\end{aligned}`.

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book,
`m(\text{book})`, was eliminated in this process.

`\begin{aligned}&\frac{\left(2.143* 10^3\right)\cdot m(\text{book})}{\left(2.943* 10^3\right) \cdot m(\text{book})}* 100\% \cr &= (2.143* 10^3)/(2.943* 10^3)* 100\% \cr & \approx 73\%\end{aligned}`.