# 0.3 Gravity and mechanical energy  (Page 8/9)

 Page 8 / 9

From this we see that when an object is lifted, like the suitcase in our example, it gains potential energy. As it falls back to the ground, it will lose this potential energy, but gain kinetic energy. We know that energy cannot be created or destroyed, but only changed from one form into another. In our example, the potential energy that the suitcase loses is changed to kinetic energy.

The suitcase will have maximum potential energy at the top of the cupboard and maximum kinetic energy at the bottom of the cupboard. Halfway down it will have half kinetic energy and half potential energy. As it moves down, the potential energy will be converted (changed) into kinetic energy until all the potential energy is gone and only kinetic energy is left. The $19,6\phantom{\rule{2pt}{0ex}}J$ of potential energy at the top will become $19,6\phantom{\rule{2pt}{0ex}}J$ of kinetic energy at the bottom.

During a flood a tree truck of mass $100\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ falls down a waterfall. The waterfall is $5\phantom{\rule{2pt}{0ex}}m$ high. If air resistance is ignored, calculate

1. the potential energy of the tree trunk at the top of the waterfall.
2. the kinetic energy of the tree trunk at the bottom of the waterfall.
3. the magnitude of the velocity of the tree trunk at the bottom of the waterfall.

• The mass of the tree trunk $m=100\phantom{\rule{2pt}{0ex}}\mathrm{kg}$
• The height of the waterfall $h=5\phantom{\rule{2pt}{0ex}}m$ . These are all in SI units so we do not have to convert.
• Potential energy at the top
• Kinetic energy at the bottom
• Velocity at the bottom
1. $\begin{array}{ccc}\hfill PE& =& mgh\hfill \\ \hfill PE& =& \left(100\right)\left(9,8\right)\left(5\right)\hfill \\ \hfill PE& =& 4900\phantom{\rule{3.33333pt}{0ex}}\mathrm{J}\hfill \end{array}$
2. The kinetic energy of the tree trunk at the bottom of the waterfall is equal to the potential energy it had at the top of the waterfall. Therefore $KE=4 900\phantom{\rule{2pt}{0ex}}J$ .

3. To calculate the velocity of the tree trunk we need to use the equation for kinetic energy.

$\begin{array}{ccc}\hfill KE& =& \frac{1}{2}m{v}^{2}\hfill \\ \hfill 4900& =& \frac{1}{2}\left(100\right)\left({v}^{2}\right)\hfill \\ \hfill 98& =& {v}^{2}\hfill \\ \hfill v& =& 9,899...\hfill \\ \hfill v& =& 9,90\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\mathrm{downwards}\hfill \end{array}$

A $2\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ metal ball is suspended from a rope. If it is released from point $A$ and swings down to the point $B$ (the bottom of its arc):

1. Show that the velocity of the ball is independent of it mass.
2. Calculate the velocity of the ball at point $B$ .

• The mass of the metal ball is $m=2\phantom{\rule{2pt}{0ex}}\mathrm{kg}$
• The change in height going from point $A$ to point $B$ is $h=0,5\phantom{\rule{2pt}{0ex}}m$
• The ball is released from point $A$ so the velocity at point, ${v}_{A}=0\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ .

All quantities are in SI units.

• Prove that the velocity is independent of mass.
• Find the velocity of the metal ball at point $B$ .
1. As there is no friction, mechanical energy is conserved. Therefore:

$\begin{array}{ccc}\hfill {U}_{A}& =& {U}_{B}\hfill \\ \hfill P{E}_{A}+K{E}_{A}& =& P{E}_{B}+K{E}_{B}\hfill \\ \hfill mg{h}_{A}+\frac{1}{2}m{\left({v}_{A}\right)}^{2}& =& mg{h}_{B}+\frac{1}{2}m{\left({v}_{B}\right)}^{2}\hfill \\ \hfill mg{h}_{A}+0& =& 0+\frac{1}{2}m{\left({v}_{B}\right)}^{2}\hfill \\ \hfill mg{h}_{A}& =& \frac{1}{2}m{\left({v}_{B}\right)}^{2}\hfill \end{array}$

As the mass of the ball $m$ appears on both sides of the equation, it can be eliminated so that the equation becomes:

$g{h}_{A}=\frac{1}{2}{\left({v}_{B}\right)}^{2}$
$2g{h}_{A}={\left({v}_{B}\right)}^{2}$

This proves that the velocity of the ball is independent of its mass. It does not matter what its mass is, it will always have the same velocity when it falls through this height.

2. We can use the equation above, or do the calculation from 'first principles':

$\begin{array}{ccc}\hfill {\left({v}_{B}\right)}^{2}& =& 2g{h}_{A}\hfill \\ \hfill {\left({v}_{B}\right)}^{2}& =& \left(2\right)\left(9.8\right)\left(0,5\right)\hfill \\ \hfill {\left({v}_{B}\right)}^{2}& =& 9,8\hfill \\ \hfill {v}_{B}& =& \sqrt{9,8}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \end{array}$

#### Questions & Answers

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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2
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