Converting joules to newtons

  1. BitWiz

    BitWiz 64
    Gold Member

    I'm having trouble with this one: How do I convert joules to newtons?

    Say I have a source of energy E that is capable of supplying 10 joules -- per second -- that I wish to use to accelerate a body. Would the energy supply then look something like this(?):

    E/s = 10 * 1kg * m^2 / s^2 /s
    = 10kg * m^2 / s^3

    If I divide this by m/s -- which seems to be (v)elocity -- I get(?):

    E/m = 10kg * m / s^2

    ie 10 newtons.

    So is that the way it works in this case? Divide E/s by the *current* (non-Lorentz) velocity to get E/m as (instantaneous) newtons? If so, what happens when the velocity is (or approaches) zero? In the expression E/m, which "m" is this?

  2. jcsd
  3. Redbelly98

    Redbelly98 12,020
    Staff Emeritus
    Science Advisor
    Homework Helper

    This calculation is not a units conversion, since energy and force are different concepts.

    Can you provide more details about the situation you are analyzing? Is a mass being accelerated from rest, and you wish to relate the applied force to the final kinetic energy of the object?

    p.s. welcome to Physics Forums.
  4. BitWiz

    BitWiz 64
    Gold Member

    Hi, Redbelly,

    Thanks for the reply.
    ... and over which I've been losing a lot of sleep. ;-) There seems to be a math relationship, but I haven't had a lot of luck finding a real-world equivalence.
    Kinetic energy is another matter I've been struggling with, so perhaps it's key. I thought I understood the concept, but ... there's a problem (more later).

    Say I want to accelerate a body in gravity-free, friction-free space using a very simple nuclear reactor that is capable of a single nuclear reaction. Say I bang a couple deuterium nuclei into one another and manage to harvest the energy in a neutron with X MeV or J. I want to use this energy (with 100% efficiency) to accelerate a mass over a period of one second. How do I calc my delta-v?

    Now say the reactor can run continuously at 10 reactions per second. I'm now getting 10X MeV/s. How does this affect my delta-v?

    Finally -- and here's the kinetic question -- if I have a body at rest of 101 kg, and I take 1kg of it and use 100 newtons to accelerate it to 100m/s in one second, the remainder of the body now has a velocity of 1 m/s, and a kinetic energy of 100J (?) If I now take another kg from the mass, and accelerate it the same way, the remaining 99 kg now has roughly double the velocity -- and momentum(?) -- but four times the kinetic energy in joules(?) -- though I only used twice the newtons to get there? <smoke coming out of ears>[/QUOTE]
    Thank you, and thank you for your kind attention. I'll try not to be a pest. ;-)

  5. Andrew Mason

    Andrew Mason 6,801
    Science Advisor
    Homework Helper

    As Redbelly points out, Joules are units for energy and Newtons are units of force.

    Energy per unit time is a measure of power and would be in units of Joules/sec or Newton-metres/sec. (Force x distance/time)

    So power can be thought of as force x speed. Dividing power by speed gives the force.

  6. BitWiz

    BitWiz 64
    Gold Member

    Thanks, Andrew. I think that's what I came up with in the opening post (but not as elegantly). So as speed approaches zero, force goes to infinity ... ? Does this have an analogy in the real world? (a stalled electric motor sucking amps?)

    Re, the original post: The output of nuclear reactions is measured in MeV or joules, and thank you for pointing out that J/sec is "power." But although they are seemingly incompatible ideas, I need to use this power to accelerate some mass, and it would simplify my setup if I could think of it in the equivalent number of newtons. I thought I had all this straight until "kinetic energy" of the accelerated mass came along, and everything blew up.

    If I accelerate a car to 30 mph using a constant wheel torque, then accelerate it to 60 mph using the same wheel torque, I know I have added three times the 30mph-kinetic energy to the total energy to come up with that of 60mph. That makes sense to me -- for one thing, the latter 30 mph pushed a lot more gas through the engine to maintain that torque.

    But if instead, I stand in a car at rest and throw a weight out the back fast enough to accelerate the car to 30 mph using Newton's Third, then repeat this with another weight to get the car to 60, somehow I've got four times the kinetic energy while only doubling the newtons. If I repeat and triple the newtons, now I have nine times the KE. I seem to be creating energy, and since that can't be, I must be missing something pretty important. Can you help me connect the dots?

  7. Andrew Mason

    Andrew Mason 6,801
    Science Advisor
    Homework Helper

    You have to be careful about the frame of reference that you are measuring KE in. You also have to take into consideration the momentum and kinetic energy of the entire system.

    When a car accelerates on the earth, it pushes back on the road. So there is momentum - and kinetic energy imparted to the earth. Because the earth is so massive, you cannot measure the change in energy of the earth, so we ignore it. Suppose instead of the car accelerating on the road, it is sitting on a barge floating freely in a lake of frictionless water. What happens to the energy delivered by the engine to the wheels? Can you just take into account the kinetic energy of the car relative to the barge? What frame of reference should you measure change of KE in when trying to relate it to the energy output of the engine?

    If you push back on something much lighter by ejecting it at high speed, you have to take into account the energy of the matter you are ejecting. How does that change as you accelerate the car? Since you are ejecting mass, what happens to the mass of the car that you are accelerating? How does that differ from the first case of the car accelerating on a road?

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