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

A car battery is rated to be able to move 600 ampere-hours (A·h) 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?


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 PE elec = qV size 12{"PE" rSub { size 8{"elec"} } = ital "qV"} {} , we have to calculate the charge q size 12{q} {} in 600 A·h , which is the product of the current I and the time t size 12{t} {} . We then multiply the result by 12.0 V. We can then calculate the battery’s increase in mass using Δ E = PE elec = ( Δ m ) c 2 size 12{ΔE="PE" rSub { size 8{"elec"} } = \( Δm \) c rSup { size 8{2} } } {} . Part (b) is a simple ratio converted to a percentage.

Solution for (a)

  1. Identify the knowns. I t = 600 A h ; V = 12 . 0 V size 12{V="12" "." 0`V} {} ; c = 3 . 00 × 10 8 m/s
  2. Identify the unknown. Δ m size 12{m} {}
  3. Choose the appropriate equation. PE elec = ( Δ m ) c 2
  4. Rearrange the equation to solve for the unknown. Δ m = PE elec c 2 size 12{m= { {"PE" rSub { size 8{"elec"} } } over {c rSup { size 8{2} } } } } {}
  5. Plug the knowns into the equation.
    Δ m = PE elec c 2 = qV c 2 = ( I t ) V c 2 = ( 600 A h ) ( 12.0 V ) ( 3.00 × 10 8 ) 2 .

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

    Δ m = ( 600 C/s h 3600 s 1 h ( 12.0 J/C ) ( 3.00 × 10 8 m/s ) 2 = ( 2.16 × 10 6 C ) ( 12.0 J/C ) ( 3.00 × 10 8 m/s ) 2

    Using the conversion 1 kg m 2 /s 2 = 1 J , we can write the mass as

    Δ m = 2.88 × 10 10 kg .

Solution for (b)

  1. Identify the knowns. Δ m = 2.88 × 10 10 kg ; m = 20.0 kg
  2. Identify the unknown. % change
  3. Choose the appropriate equation. % increase = Δ m m × 100% size 12{%" increase"= { {Δm} over {m} } times "100"%} {}
  4. Plug the knowns into the equation.
    % increase = Δ m m × 100% = 2.88 × 10 10 kg 20.0 kg × 100% = 1.44 × 10 9 % .


Both the actual increase in mass and the percent increase are very small, since energy is divided by c 2 size 12{c rSup { size 8{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 1 2 mv 2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{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 net = KE . size 12{W rSub { size 8{"net"} } ="KE"} {}

Relativistically, at rest we have rest energy E 0 = mc 2 . The work increases this to the total energy E = γmc 2 . Thus,

Questions & Answers

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Aliyu Reply
what kind of error do you think? and work is held by which force?
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Douglas Reply
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Emmanuel Reply
so what question are you passing across... sir?
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Emmanuel Reply
54 joule
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
i dont think dere is any potential energy... by d virtue of no height present
there is compressed energy,dats only potential energy na?
yes.. but... how will u approach that question without The Height in the question?
Can you explain how you get 54J?
Because mine is 36J
got 36J too
OK the answer is 54J Babar is correct
Conservation of Momentum
woow i see.. can you give the formula for this
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 Reply
Please help!
please help find dy/dx 2x-y/x+y
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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
i think i m correct
But how?
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kwame Reply
yes what is it?
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it is a theory on how the universe began. to understand more I would suggest researching the topic online.
thanks guys
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Joshua Reply
We can seek accelation first
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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
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pls í need help
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now use this formula
what's the answer then
great Mudang
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hey mudang there is a product of force and acceleration not force and displacement
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its not possible
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isidor Reply
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technical Reply
if i am going to start studying physics where should i start?
I think from kinematics
You can find physics books at the library or online. That's how I started.
And yes, kinematics is usually where you can begin.
study basic algebra and calculus and can start from classical mechanics
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benjamin Reply
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any methods can take to solve this eqtions
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what is the question dear
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find the distance
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?
Hayne Reply
How can we calculate without any information?
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Practice Key Terms 3

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