# 0.3 Gravity and mechanical energy  (Page 6/9)

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$KE=\frac{1}{2}m{v}^{2}$

Consider the $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ suitcase on the cupboard that was discussed earlier. When the suitcase falls, it will gain velocity (fall faster), until it reaches the ground with a maximum velocity. The suitcase will not have any kinetic energy when it is on top of the cupboard because it is not moving. Once it starts to fall it will gain kinetic energy, because it gains velocity. Its kinetic energy will increase until it is a maximum when the suitcase reaches the ground.

A $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ brick falls off a $4\phantom{\rule{2pt}{0ex}}m$ high roof. It reaches the ground with a velocity of $8,85\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ . What is the kinetic energy of the brick when it starts to fall and when it reaches the ground?

• The mass of the rock $m=1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$
• The velocity of the rock at the bottom ${v}_{\mathrm{bottom}}=8,85\phantom{\rule{2pt}{0ex}}m·$ s ${}^{-1}$

These are both in the correct units so we do not have to worry about unit conversions.

1. We are asked to find the kinetic energy of the brick at the top and the bottom. From the definition we know that to work out $KE$ , we need to know the mass and the velocity of the object and we are given both of these values.

2. Since the brick is not moving at the top, its kinetic energy is zero.

3. $\begin{array}{ccc}\hfill KE& =& \frac{1}{2}m{v}^{2}\hfill \\ & =& \frac{1}{2}\left(1\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right){\left(8,85\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)}^{2}\hfill \\ & =& 39,2\phantom{\rule{0.166667em}{0ex}}\mathrm{J}\hfill \end{array}$

## Checking units

According to the equation for kinetic energy, the unit should be $\mathrm{kg}·m{}^{2}·s{}^{-2}$ . We can prove that this unit is equal to the joule, the unit for energy.

$\begin{array}{ccc}\hfill \left(\mathrm{kg}\right){\left(\mathrm{m}·{\mathrm{s}}^{-1}\right)}^{2}& =& \left(\mathrm{kg}·\mathrm{m}·{\mathrm{s}}^{-2}\right)·\mathrm{m}\hfill \\ & =& \phantom{\rule{0.166667em}{0ex}}\mathrm{N}·\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(\mathrm{because}\mathrm{Force}\left(\mathrm{N}\right)=\mathrm{mass}\left(\mathrm{kg}\right)×\mathrm{acceleration}\left(\mathrm{m}·{\mathrm{s}}^{-2}\right)\right)\hfill \\ & =& \phantom{\rule{0.166667em}{0ex}}\mathrm{J}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(\mathrm{Work}\left(\mathrm{J}\right)=\mathrm{Force}\left(\mathrm{N}\right)×\mathrm{distance}\left(\mathrm{m}\right)\right)\hfill \end{array}$

We can do the same to prove that the unit for potential energy is equal to the joule:

$\begin{array}{ccc}\hfill \left(\mathrm{kg}\right)\left(\mathrm{m}·{\mathrm{s}}^{-2}\right)\left(\mathrm{m}\right)& =& \phantom{\rule{0.166667em}{0ex}}\mathrm{N}·\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ & =& \phantom{\rule{0.166667em}{0ex}}\mathrm{J}\hfill \end{array}$

A bullet, having a mass of $150\phantom{\rule{2pt}{0ex}}g$ , is shot with a muzzle velocity of $960\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ . Calculate its kinetic energy.

• We are given the mass of the bullet $m=150\phantom{\rule{2pt}{0ex}}g$ . This is not the unit we want mass to be in. We need to convert to kg.
$\begin{array}{ccc}\hfill \mathrm{Mass}\phantom{\rule{3.33333pt}{0ex}}\mathrm{in}\phantom{\rule{3.33333pt}{0ex}}\mathrm{grams}÷1000& =& \mathrm{Mass}\phantom{\rule{3.33333pt}{0ex}}\mathrm{in}\phantom{\rule{3.33333pt}{0ex}}\mathrm{kg}\hfill \\ \hfill 150\phantom{\rule{3.33333pt}{0ex}}g÷1000& =& 0,150\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\hfill \end{array}$
• We are given the initial velocity with which the bullet leaves the barrel, called the muzzle velocity, and it is $v=960\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ .
• We are asked to find the kinetic energy.
1. We just substitute the mass and velocity (which are known) into the equation for kinetic energy:

$\begin{array}{ccc}\hfill KE& =& \frac{1}{2}m{v}^{2}\hfill \\ & =& \frac{1}{2}\left(0,150\right){\left(960\right)}^{2}\hfill \\ & =& 69\phantom{\rule{0.166667em}{0ex}}120\phantom{\rule{0.166667em}{0ex}}\mathrm{J}\hfill \end{array}$

## Kinetic energy

1. Describe the relationship between an object's kinetic energy and its:
1. mass and
2. velocity
2. A stone with a mass of $100\phantom{\rule{2pt}{0ex}}g$ is thrown up into the air. It has an initial velocity of $3\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ . Calculate its kinetic energy
1. as it leaves the thrower's hand.
2. when it reaches its turning point.
3. A car with a mass of $700\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ is travelling at a constant velocity of $100\phantom{\rule{2pt}{0ex}}\mathrm{km}·\mathrm{hr}{}^{-1}$ . Calculate the kinetic energy of the car.

## Mechanical energy

Mechanical energy is the sum of the gravitational potential energy and the kinetic energy.

Mechanical energy, $U$ , is simply the sum of gravitational potential energy ( $PE$ ) and the kinetic energy ( $KE$ ). Mechanical energy is defined as:

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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silver nanoparticles could handle the job?
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