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How is work measured in a vacuum?

  1. Jun 21, 2015 #1
    I've just learned that work is done when a force moves something, and it's measured in joules.
    I've also learned that work = force x distance, but what significance does distance have? The force itself is what's being measured.

    In other words, if an object weighing 1kg was accelerated to 5m/s2 in a vacuum, what work have I applied on the object? Its distance is essentially infinite, because it keeps traveling without resistance.

    Also, if a 1kg object on earth is lifted (with a force of 9.8N), how does the distance make any difference? The force applied is still the same, regardless of how far the object travels. What exactly is 'work' measuring?

    Could someone please clarify how this works? Thanks for your help.
     
  2. jcsd
  3. Jun 21, 2015 #2

    ZapperZ

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    Did you forget Newton's first law and the presence of inertia?

    An object will either remain stationary, or move with a constant velocity unless a net force acts on it. You need to apply a force to cause something to accelerate, and this acceleration changes the object's velocity. Thus, if you are pushing it with your hand, you will feel a "resistance" back from the object. This is where you have to expand energy to cause the object to accelerate.

    If you have trouble understanding this, do the reverse. Try to slow down to a complete stop an object that is already in motion. Don't you think you have to exert a force to slow it down? Don't you feel the object pushing back on you when you do this? This is the same thing, except in reverse!

    Zz.
     
  4. Jun 21, 2015 #3
    Traveling x more distance = x more work done.
     
  5. Jun 22, 2015 #4
    Physicists and engineers would say "work" has been done on an object if its energy has been changed. That could be its kinetic energy, its gravitational potential, its heat energy .... So for your 1 kg object in space; if its speed has changed by 5 m/s (note: you said "accelerated to 5m/s2" which doesn't quite make sense.) then its kinetic energy has changed by 12.5 Joules (one half mass times velocity squared). So we would say 12.5Joules of work has been done on that mass. If you follow through with the mathematics you will discover that force required to create the change in kinetic energy multiplied by the distance that the mass travels while the force is being applied is also equal to the work done. For this example a force of 12.5 Newtons applied while the object moved a metre would get its velocity up to 5 m/s or a force of 1 Newton applied for 12.5 metres would achieve the same result. The distance traveled by the object after the force ceases is irrelevant.

    For an object of 1 kg near the Earth's surface applying a force of 9.8 Newtons vertically upwards will not do any work. The force of gravity will be pulling downwards with exactly the same force and the object will remain stationary. If there is nothing acting except gravity, the object will accelerate downwards at 9.8 m/s2. After one second the object will be traveling at 9.8 m/s and have a kinetic energy of about 48 Joules. So 48 joules of work will have been done on it by gravity. If you then use W = f x d to calculate d you find that after one second the object has fallen 4.9 metres. Note that energy is conserved so it will have lost 48 J of gravitational potential energy. Also note that the "force" of gravity is not 9.8 Newtons. The acceleration due to gravity is 9.8 m/s2 near the Earth's surface. On a one kilogram mass this means it exerts 9.8 Newtons but that is only because we chose a mass of 1 kg.

    Does that help?
     
  6. Jun 22, 2015 #5
    but it is relevant before that.
     
  7. Jun 24, 2015 #6
    You do not apply work on anything actually, you apply force. Here the force is equal to F=ma

    With regarding to work you should consider when it travels x diplacement
    Work, W=Fx
    When it travels x' displacement
    Work W=Fx'

    It all depends on how much displacement occured from its initial position. The more displacement this case the more work will be done,

    For example in your qurstion the force is 5*1=5N

    So, when the body will cover a displacement 2m hen work will be 5*2=10J,
    When 3 m then W=5*3=15 J

    That is, with time it will cover more distance and more work will be done until it is stopped by some other forces. (That is actually the work done by the force you applied, here 5N)
     
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