An earthquake causes a 3 kg book to fall from a shelf. If the book lands with a speed of 4 m/s, from what height did it fall?

Mar 10, 2017

$\approx 0.82 m$

Explanation:

This can be solved using either kinematics or energy conservation, both methods assuming no outside forces.

Energy Conservation

$\Delta E = 0$

$\implies {E}_{i} = {E}_{f}$

$\implies {U}_{i} + {K}_{i} = {U}_{f} + {K}_{f}$

Where $U$ is the potential energy and $K$ is the kinetic energy.

Initially, the book has only gravitational potential energy, as it is stationary above the surface of the earth and $\vec{v} = 0$. Just before the book hits the ground, all of the gravitational potential energy it had originally has been converted into kinetic energy, i.e. it has only kinetic energy finally. Our statement for energy conservation now becomes:

${U}_{g} = K$

Where ${U}_{g} = m g h$ and $K = \frac{1}{2} m {v}^{2}$.

$\implies m g h = \frac{1}{2} m {v}^{2}$

We can solve for $h$, the height that the book fell from.

$h = \frac{\frac{1}{2} \cancel{m} {v}^{2}}{\cancel{m} g}$

$\implies h = {v}^{2} / \left(2 g\right)$

We are given that $v = 4 \frac{m}{s}$, and we know that $g = 9.8 \frac{m}{s} ^ 2$.

$h = {\left(4 \frac{m}{s}\right)}^{2} / \left(2 \cdot 9.8 \frac{m}{s} ^ 2\right)$

$\approx 0.82 m$

$\therefore$The book fell from a height $\approx 0.82 m$

Kinematics

${v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta y$

Because the book begins from rest:

$0 = {v}_{i}^{2} + 2 a \Delta y$

Now we solve for $y$ with a bit of algebra:

$\implies y = \frac{- {v}_{i}^{2}}{2 a}$

We plug in our values:

$\implies y = \frac{- {\left(4 \frac{m}{s}\right)}^{2}}{2 \cdot - 9.8 \frac{m}{s} ^ 2}$

$\implies y \approx 0.82 m$.

Note that $9.8 \frac{m}{s} ^ 2$ is positive when we use energy conservation because we are using $g$, which is a constant that is always positive. We use $- 9.8 \frac{m}{s} ^ 2$ when we use kinematics because for an object in free-fall, the acceleration is equal to $- g$.