# 0.1 Gravity and mechanical energy  (Page 5/5)

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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 $h=0,5\phantom{\rule{2pt}{0ex}}\mathrm{m}$
• The ball is released from point $A$ so the velocity at point, ${v}_{A}=0\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ .

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:

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

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':

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

## Potential energy

1. A tennis ball, of mass $120\phantom{\rule{2pt}{0ex}}\mathrm{g}$ , is dropped from a height of $5\phantom{\rule{2pt}{0ex}}\mathrm{m}$ . Ignore air friction.
1. What is the potential energy of the ball when it has fallen $3\phantom{\rule{2pt}{0ex}}\mathrm{m}$ ?
2. What is the velocity of the ball when it hits the ground?
2. A bullet, mass $50\phantom{\rule{2pt}{0ex}}\mathrm{g}$ , is shot vertically up in the air with a muzzle velocity of $200\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ . Use the Principle of Conservation of Mechanical Energy to determine the height that the bullet will reach. Ignore air friction.
3. A skier, mass $50\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ , is at the top of a $6,4\phantom{\rule{2pt}{0ex}}\mathrm{m}$ ski slope.
1. Determine the maximum velocity that she can reach when she skies to the bottom of the slope.
2. Do you think that she will reach this velocity? Why/Why not?
4. A pendulum bob of mass $1,5\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ , swings from a height A to the bottom of its arc at B. The velocity of the bob at B is $4\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ . Calculate the height A from which the bob was released. Ignore the effects of air friction.
5. Prove that the velocity of an object, in free fall, in a closed system, is independent of its mass.

## Energy graphs - (not in caps, included for completeness)

Let us consider our example of the suitcase on the cupboard, once more.

Let's look at each of these quantities and draw a graph for each. We will look at how each quantity changes as the suitcase falls from the top to the bottom of the cupboard.

• Potential energy : The potential energy starts off at a maximum and decreases until it reaches zero at the bottom of the cupboard. It had fallen a distance of 2 metres.
• Kinetic energy : The kinetic energy is zero at the start of the fall. When the suitcase reaches the ground, the kinetic energy is a maximum. We also use distance on the $x$ -axis.
• Mechanical energy : The mechanical energy is constant throughout the motion and is always a maximum. At any point in time, when we add the potential energy and the kinetic energy, we will get the same number.

## Summary

• The potential energy of an object is the energy the object has due to his position above a reference point.
• The kinetic energy of an object is the energy the object has due to its motion.
• Mechanical energy of an object is the sum of the potential energy and kinetic energy of the object.
• The unit for energy is the joule (J).
• The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be changed from one form into another.
• The Law of Conservation of Mechanical Energy states that the total mechanical energy of an isolated system remains constant.
• The table below summarises the most important equations:
 Potential Energy ${E}_{P}=mgh$ Kinetic Energy ${E}_{K}=\frac{1}{2}m{v}^{2}$ Mechanical Energy ${E}_{M}={E}_{K}+{E}_{P}$

## End of chapter exercises: gravity and mechanical energy

1. Give one word/term for the following descriptions.
1. The force with which the Earth attracts a body.
2. The unit for energy.
3. The movement of a body in the Earth's gravitational field when no other forces act on it.
4. The sum of the potential and kinetic energy of a body.
5. The amount of matter an object is made up of.
2. Consider the situation where an apple falls from a tree. Indicate whether the following statements regarding this situation are TRUE or FALSE. Write only 'true' or 'false'. If the statement is false, write down the correct statement.
1. The potential energy of the apple is a maximum when the apple lands on the ground.
2. The kinetic energy remains constant throughout the motion.
3. To calculate the potential energy of the apple we need the mass of the apple and the height of the tree.
4. The mechanical energy is a maximum only at the beginning of the motion.
5. The apple falls at an acceleration of $9,8\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-2}$ .
3. A man fires a rock out of a slingshot directly upward. The rock has an initial velocity of $15\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ .
1. What is the maximum height that the rock will reach?
2. Draw graphs to show how the potential energy, kinetic energy and mechanical energy of the rock changes as it moves to its highest point.
4. A metal ball of mass $200\phantom{\rule{2pt}{0ex}}\mathrm{g}$ is tied to a light string to make a pendulum. The ball is pulled to the side to a height (A), $10\phantom{\rule{2pt}{0ex}}\mathrm{cm}$ above the lowest point of the swing (B). Air friction and the mass of the string can be ignored. The ball is let go to swing freely.
1. Calculate the potential energy of the ball at point A.
2. Calculate the kinetic energy of the ball at point B.
3. What is the maximum velocity that the ball will reach during its motion?
5. A truck of mass $1,2\phantom{\rule{2pt}{0ex}}\mathrm{tons}$ is parked at the top of a hill, $150\phantom{\rule{2pt}{0ex}}\mathrm{m}$ high. The truck driver lets the truck run freely down the hill to the bottom.
1. What is the maximum velocity that the truck can achieve at the bottom of the hill?
2. Will the truck achieve this velocity? Why/why not?
6. A stone is dropped from a window, $6\phantom{\rule{2pt}{0ex}}\mathrm{m}$ above the ground. The mass of the stone is $25\phantom{\rule{2pt}{0ex}}\mathrm{g}$ . Use the Principle of Conservation of Energy to determine the speed with which the stone strikes the ground.

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what is the actual application of fullerenes nowadays?
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or in general
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in general
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