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Homework Help: The definition of 1 Joule

  1. Oct 19, 2006 #1
    Someone told me that 1 Joule is roughly the amount of energy required to lift a 1 kilo object 10 cm of the ground. This sounds wierd to me. Seems the amount of energy required would depend on how fast you lifted the object? The more time you were to spend lifting, the more energy you would spend fighting gravity....or? If someone could just clarify this to me I would be greatful :)
  2. jcsd
  3. Oct 19, 2006 #2


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    No, that's not true at all. The energy you use to lift something is the force needed (the weight of the object) times the distance. It doesn't matter how long you take to lift it.

    You need to understand that the "energy" in this sense is not necessarily the energy you might burn in your muscles since there is a lot from the small muscles pulling against each other, etc. In particular, it does not take any energy at all to hold a heavy object at a constant height off the ground, any more than a table uses energy to hold an object there. The tiredness you feel after trying to hold up a heavy object for a long time is not energy imparted to the object but the result of the fact that your muscles are not 100% efficient and is not counted in mechanics.

    Actually, the " energy required to lift a 1 kilo object 10 cm of the ground" is a good approximation to a Joule but not exactly true. Two equivalent definitions of the "Joule" are "the work done by a 1 Newton force in moving an object 1 meter" and "the kinetic energy of a 1 kg object moving at 1 meter per second". Since a Newton is defined as "the force necessary to accelerate a 1 kg object at 1 meter per second per second", it has basic units of kg m/s2 and so, multiplying by that "1 meter", a Joule has basic units of kg m2/s2. Kinetic energy, on the other hand is (1/2)mv2 and so has basic units of "kg times (meter per second squared)" or kg(m2/s2 just as before.

    Since the acceleration due to gravity at the surface of the earth is about 9.8 m/s2, the weight of a 1 kg object is 1(9.8)= 9.8 Newtons= 9.8 kg m/s2 and so lifting it 10 cm= 0.1 m you have done work (imparted energy to the object) of (0.1 cm)(9.8 kg m/s2)= 0.98 kg m2/s2= 0.98 Joules, not exactly one Joule.
  4. Oct 19, 2006 #3
    Hey, thx for replying! Still I'm a bit confused...what do you mean by that the energy my muscles use is not "imparted to the object"? And that they aren't 100 % efficient?
  5. Oct 19, 2006 #4
    Take some time and think about what happens when you lift something. Say you are deadlifting a barbell (meaning you squat down and pick up the weight to your knees using just your legs). Your arms don't move, but yet they start to fatigue as you stand there with the weight hanging down at your knees. Your muscles are doing many things standing there. One thing is the contracted muscle has to stretch the opposite muscle. Your heart might have to work harder to push blood by the vessels being squished by the weight. Your cells are repairing themselves. etc... So your body still has lots to do just standing there -- I mean you still burn calories even when you sleep!

    So, the energy your muscles are using is not moving the weight anywhere, but the body has to work to keep itself going.

    As far as efficiency -- in any chemical reaction you are going to lose some energy (if none is added to the system). It usually is in the form of heat, but can be chemical, electrical, etc. So not all of your energy can be put into your deadlift, some gets lost.
  6. Oct 19, 2006 #5
    One joule is the amount of energy required to exert a force of 1N through a distance of 1 meter. It doesn't matter how fast you do this. The RATE at which you use this energy is power. A watt is a measure of power, which is a joule/sec.

    Energy is the capacity to do work, no matter what form it is in, whether it is electrical energy, mechanical energy, chemical energy, heat, etc. 5 joules of energy, no matter what form, can do the same amount of work. You can interconvert energies between different types. If there is a net work acting on an object then you will accelerate or decellerate the object and change its kinetic energy.
    Last edited: Oct 19, 2006
  7. Sep 3, 2007 #6
    Hi, I would like to know, what is the difference between a 7 ton Harrier placed on a 1m high table, and the same 7 ton Harrier, maintaining an stationary altitude of 1 meter above the ground with its vertical thrusters ???

    I both the cases, the potential energy of the Harrier is the same.

    Why is maintaining the height of 1 meter, without a table to provide the upward-normal, so energy consuming ???

    I believe the concept applies to holding a barbell, at 1m height and getting tired over time.

    How do we calculate the energy required for maintenance of height ????
    Last edited: Sep 3, 2007
  8. Sep 3, 2007 #7


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    On a table, the Harrier is supported by the table. Without external support, the Harrier must use thrust from it's engines, and that requires power. The integral of power over time yields energy.

    No. This is a different situation. Muscles under load (stress) must expend energy (from metabolism). The metabolism is anaerobic when one is holding a barbell continually.

    The integral of power over time yields energy.
  9. Sep 3, 2007 #8
    Does that mean that the energy spent by Superman to hold a Harrier at 1m & the energy the Harrier's thrusters consume at 1m is not the same ?

    I believe, how Superman holds the Harrier is important.
    If he places on it on his head, and makes sure that the ground's normal, goes through this leg bones & spine and head, he would consume little energy.
    (Energy spent would be equal to the energy required by the muscles to keep the bones in line)

    But let assume he holds it like a person giving offering,

    Code (Text):

    i      /
    i      \
    Won't the energy spent by him be at least close to what the Harrier would consume on its thrusters ???
    Last edited: Sep 3, 2007
  10. Sep 3, 2007 #9


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    Why does one invoke a fictional character in this problem? Superman seems to 'fly' with no evident propulsion system, which contradicts physical laws as we know/understand them.

    Putting a mass on a table is a 'static' problem. The table is under compression due to the weight imposed by the mass on the table in a gravitational field.

    Without external support, the Harrier must use 'thrust' - a force - to stay aloft. Thrust requires mass flow, which requires work/energy from its engines. This is a 'dynamic' problem.
  11. Jun 2, 2010 #10
    I would like to screw things up a little bit. Here is a little thought experiment that I want to share. But first, the definition of 1 Joule:

    1 Joule = 1 Newton * 1 meter

    Now imagine three objects, A, B and C, all with the mass 1 kg, floating around in space. A and B moves uniformly relative to C, with the speed of 1 m/s.

    Now, A has a rocket engine, and a fuel tank. It contains exactly enough fuel to apply the force 1N over 1m relative to B(!!!). My question is, how much energy does the fuel contain?

    The easy answer is 1 Joule (relative to B). But if you measure it relative to C, things get quite interesting. The time it takes to accelerate and travel 1m relative to B is found as follows:

    1m = acceleration * t2 / 2 =>
    t = sqrt(2) = 1.4142...

    How far does A travel relative to C during this time?

    distance = vt + at2 / 2

    ...and we have t = 1.4142..., v = 1m/s and a = 1m/s2, which gives a distance of 2.4142 meters. That means that we have applied the force 1N over more than 2 meters, which in turn means that we have spent 2.4142 Joules of energy (relative to C)! This is inconsistent, and means that the engine's effect increases as the speed of A increases. So something must be wrong. But what?
    Last edited: Jun 2, 2010
  12. Mar 22, 2012 #11
    To Nizzeberra,
    There is nothing inconsistent in your problem. You said A and B are already moving at 1m/s relative to C. So in 1.4142 seconds, they will have used 1.4142 J from their current state of motion. The balance, 1J, comes from the rocket engine for a total of 2.4142.
  13. Mar 23, 2012 #12


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    Hi sitichon. Welcome to physicsforums!

    The thread you have replied to here was last active in 2010. The person you have replied to hasn't logged in since May of last year, so it is likely he may no longer be following the thread, or at least by now may have his answer. https://www.physicsforums.com/images/icons/icon6.gif [Broken]

    To decide where your contributions are likely to be appreciated on physics forums, take note of the date at the top of the last post in a thread. If it is more than a few months old, your efforts there may be to no avail. There are plenty of recent threads where assistance is certain to be appreciated. :cool:

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    Last edited by a moderator: May 5, 2017
  14. May 30, 2013 #13
    I think a slight misunderstanding might have arisen from thinking that a man struggling to hold up a weight and a table holding up a weight are not doing work equally. In the presence of gravity, the table does this very sneaky like. The leg is under compression. This means that, even though the wooden leg is a solid, the molecules get a teeny tiny bit closer to each other and the bonding forces convey the energy down the leg and to the ground and eventually through the core of the earth to the OTHER side of the planet where gravity pushes back. And in that place on the other side of the planet, things are sitting on tables there too. (Obviously a table is not required but you get the idea.) So it's not like it doesn't do work, it's just in equilibrium. That's how we get the generally balanced ball shape of the planet.

    If you prefer to not directly push on a solid object, you can push on the air. And just because it's tough to see all the downward forces on those air molecules eventually hitting the ground by way of other air molecules doesn't mean they don't balance as well.

    Another interesting property of solid matter is that the zone of influence by the electron cloud around atoms is exponential in force, so in solid objects differing amounts of "weight" placed on them (like the table leg reference) actually does change the spacing of the component atoms, it's just hard to measure in very ridged materials. I'm sure if you could measure the distance across 10 billion atoms of iron at the core of the planet compared to 10 billion atoms of iron at the surface (arranged identically) you would find the surface iron to be slightly longer. Yes, virtually impossible to measure, but there. There enough to store and transmit the balancing forces.

    Furthermore, purests will tell you no work is done because nothing moves, and that is true, but that's also like saying you didn't break a sweat in a tie at a tug of war challenge.
    Last edited: May 30, 2013
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