# 28.6 Relativistic energy  (Page 3/12)

 Page 3 / 12

## Calculating rest mass: a small mass increase due to energy input

A car battery is rated to be able to move 600 ampere-hours $\left(A·h\right)$ of charge at 12.0 V. (a) Calculate the increase in rest mass of such a battery when it is taken from being fully depleted to being fully charged. (b) What percent increase is this, given the battery’s mass is 20.0 kg?

Strategy

In part (a), we first must find the energy stored in the battery, which equals what the battery can supply in the form of electrical potential energy. Since ${\text{PE}}_{\text{elec}}=\text{qV}$ , we have to calculate the charge $q$ in $600\phantom{\rule{0.25em}{0ex}}A·h$ , which is the product of the current $I$ and the time $t$ . We then multiply the result by 12.0 V. We can then calculate the battery’s increase in mass using $\Delta E={\text{PE}}_{\text{elec}}=\left(\Delta m\right){c}^{2}$ . Part (b) is a simple ratio converted to a percentage.

Solution for (a)

1. Identify the knowns. $I\cdot t=\text{600 A}\cdot \text{h}$ ; $V=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}$ ; $c=3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}$
2. Identify the unknown. $\Delta m$
3. Choose the appropriate equation. ${\text{PE}}_{\text{elec}}=\left(\Delta m\right){c}^{2}$
4. Rearrange the equation to solve for the unknown. $\Delta m=\frac{{\text{PE}}_{\text{elec}}}{{c}^{2}}$
5. Plug the knowns into the equation.
$\begin{array}{lll}\Delta m& =& \frac{{\text{PE}}_{\text{elec}}}{{c}^{2}}\\ & =& \frac{\text{qV}}{{c}^{2}}\\ & =& \frac{\left(It\right)V}{{c}^{2}}\\ & =& \frac{\left(\text{600 A}\cdot \text{h}\right)\left(\text{12.0 V}\right)}{\left(3.00×{\text{10}}^{8}{\right)}^{2}}\text{.}\end{array}$

Write amperes A as coulombs per second (C/s), and convert hours to seconds.

$\begin{array}{lll}\Delta m& =& \frac{\left(\text{600 C/s}\cdot \text{h}\left(\frac{\text{3600 s}}{\text{1 h}}\right)\left(\text{12.0 J/C}\right)}{\left(3.00×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}}\\ & =& \frac{\left(2.16×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{C}\right)\left(\text{12.0 J/C}\right)}{\left(3.00×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}}\end{array}$

Using the conversion $1\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}{\text{/s}}^{2}=1\phantom{\rule{0.25em}{0ex}}\text{J}$ , we can write the mass as

$\Delta m=2.88×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{kg}\text{.}$

Solution for (b)

1. Identify the knowns. $\Delta m=2.88×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ ; $m=\text{20.0 kg}$
2. Identify the unknown. % change
3. Choose the appropriate equation. $\text{}\text{% increase}=\frac{\Delta m}{m}×\text{100%}\text{}$
4. Plug the knowns into the equation.
$\begin{array}{lll}\text{% increase}& =& \frac{\Delta m}{m}×\text{100%}\\ & =& \frac{2.88×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{kg}}{\text{20.0 kg}}×\text{100%}\\ & =& 1.44×{\text{10}}^{-9}\text{%}\text{.}\end{array}$

Discussion

Both the actual increase in mass and the percent increase are very small, since energy is divided by ${c}^{2}$ , a very large number. We would have to be able to measure the mass of the battery to a precision of a billionth of a percent, or 1 part in ${10}^{11}$ , to notice this increase. It is no wonder that the mass variation is not readily observed. In fact, this change in mass is so small that we may question how you could verify it is real. The answer is found in nuclear processes in which the percentage of mass destroyed is large enough to be measured. The mass of the fuel of a nuclear reactor, for example, is measurably smaller when its energy has been used. In that case, stored energy has been released (converted mostly to heat and electricity) and the rest mass has decreased. This is also the case when you use the energy stored in a battery, except that the stored energy is much greater in nuclear processes, making the change in mass measurable in practice as well as in theory.

## Kinetic energy and the ultimate speed limit

Kinetic energy is energy of motion. Classically, kinetic energy has the familiar expression $\frac{1}{2}{\mathrm{mv}}^{2}$ . The relativistic expression for kinetic energy is obtained from the work-energy theorem. This theorem states that the net work on a system goes into kinetic energy. If our system starts from rest, then the work-energy theorem is

${W}_{\text{net}}=\text{KE}.$

Relativistically, at rest we have rest energy ${E}_{0}={\mathrm{mc}}^{2}$ . The work increases this to the total energy $E={\mathrm{\gamma mc}}^{2}$ . Thus,

Why is there no 2nd harmonic in the classical electron orbit?
how to reform magnet after been demagneted
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
what is the error during taking work done of a body..
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
Julius
great Mudang
Kossi
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
ancient greek language physis = nature
isidor
what is phyacs
if i am going to start studying physics where should i start?
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir