I understand that the formula for kinetic energy is: $(1/2)mv^2$, which is equal to 1 Joule or 1 N-m. If I want $m$ to be my weight (say 150lbs) and $v$ to be 40mph how do I apply it to this formula?

It would not seem like you could just plug any number (150lbs) unrelated to another number (40mph) and get back N-m.

Ignore my ignorance, I never learned physics and am trying to apply it to real world situations now.

  • $\begingroup$ lb is a measure of weight, not mass... So you must use mass in kilograms... $\endgroup$ – TanMath Oct 23 '15 at 20:40
  • $\begingroup$ en.wikipedia.org/wiki/Pound_(mass) $\endgroup$ – Mark Eichenlaub Oct 23 '15 at 20:49
  • $\begingroup$ @MarkEichenlaub but he said "If I want m to be my weight", and m is supposed to be mass.. plus, in most physics courses, and in physics applications, pound is a measure of force. That is the convention. $\endgroup$ – TanMath Oct 23 '15 at 21:01

Let's take a more familiar example.

If the cost of a baseball is $c$, the cost of ten baseballs is $10 c$.

This formula is not about dollars, euros, rupees, etc. It's a formula about cost, and it's true no matter what type of money you're talking about. The units you get out of the formula will depend on what you put in. For example, if you put in $c = \$2 $, you will get out $ \$20$. If you put in $c = 200$ yen, you will get out $2000$ yen, etc.

Similarly, the formula for kinetic energy is $\frac12 mv^2$. This is not a formula about Joules, just energy. Energy can be measured in many different units; Joules are just one of them.

If you input the mass and velocity in kilograms and meters, you will get kilograms * meters^2/seconds^2 out of it. That unit happens to be called "Joules".

If you input the mass in pounds and miles per hour, you will get pounds * miles^2/hour^2 out of it. That unit doesn't have a special name that I know of. It's still energy, though, and still uses the same formula.

You can convert pounds*miles^2/hour^2 into Joules by simply converting all the units; kilograms into pounds, miles^2 into meters^2, etc.

  • $\begingroup$ That is an excellent example. Would I be correct in stating the following? 40mph == 17.8816 m/s and 150lbs == 90.7185 kg so therefore (1/2) * 90.7185 * 17.8816 ^2 = 14503.69360416768 Joules $\endgroup$ – Johnston Oct 23 '15 at 20:58
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    $\begingroup$ that's the idea, but I don't think you converted to kg correctly $\endgroup$ – Mark Eichenlaub Oct 23 '15 at 21:08
  • $\begingroup$ OP was asking for applications of KE, which you haven't included $\endgroup$ – TanMath Oct 23 '15 at 21:29
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    $\begingroup$ @TanMath: OP was looking for how one can use different units in the KE equation (validity, possibility, etc), not applications thereof. $\endgroup$ – Kyle Kanos Oct 23 '15 at 21:37

I would first like to say that this equation is used in many different cases, and that this question is really broad. However, I do know one particular application of KE to motion.

First, you can't use lbs in the equation. It seems like you do not know the difference between mass and weight. Weight is the force of gravity pulling on you downwards. Mass is the measure of how much there is of something. Weight in Newtons can be calculated as $W=mg$ where W is the weight, m is the mass in kilograms, and g is the acceleration of gravity, which is 9.80 m/s on Earth, but is different on different planets. Therefore, weight is a variable quantity depending on where you are located (like whether you are on the Moon or on Earth), while mass is not variable.

One application of KE is in the work-energy principle. The work-energy principle states that the amount of work done on an object is equal to the change in kinetic energy. Work is defined as force through a distance - $W=Fd$ or for 2-D, $W=\vec F\cdot \vec s$. Mathematically, the work-energy principle can be stated as: $$W = \Delta K$$ or $$W=(1/2 mv_2^2)-(1/2mv_1^2) $$

If you know the forces acting upon an object along with the mass (although many times the mass is not needed as it cancels out) and the distance it travels, you can find its velocity. Or, given the necessary information, you can find the distance it travelled.


It seems I have misunderstood your question. First of all, it seems that there is a debate on whether or not pound is to be considered a mass. But for scientific purposes, pound is a force.

Second of all, you can put whatever units you want in the equation, as long as $m$ is in units of mass and $v$ is in units of velocity. You will still get an answer that is an energy. But to get units of Joules, the mass must be in kg, and the velocity must be in m/s.

  • $\begingroup$ "[pound] was originally defined as weight." Pounds were originally established when no one made a distinction between mass and weight, and since then has been effectively defined as mass for commercial purposes and force for scientific ones. It is, of course, pedagogically sound to make a strong distinction, but it doesn't have a historical basis. $\endgroup$ – dmckee Oct 23 '15 at 22:30
  • $\begingroup$ @dmckee ok.. I will edit to say for scientific purposes, pound is a force... $\endgroup$ – TanMath Oct 23 '15 at 22:35

You have to write the mass in kilograms and the velocity in meters/second to get joules. A Joule is just a kilogram, times (m/s)^2. Entering in arbitrary units to the Kinect energy formula will always get you an energy, though it will be in arbitrary energy units. A joule is just the energy of a 1 kg mass, moving at a speed of 1 m/s. If you measure mass in pounds, and velocity in miles per hour, you will still get energy, but it will be in units of energy such that a 1 pound mass is moving at 1 mile per hour... The name of that unit of energy is arbitrary.

  • $\begingroup$ lb is a unit of force so you can't use lb in the KE equation $\endgroup$ – TanMath Oct 23 '15 at 20:47
  • $\begingroup$ I know, it should be slugs, but for practical purposes I can say that a mass will read off 1 lb if I put it on a scale in an inertial frame on earth. That mass, whatever it is, will always weigh a pound if I weigh it in an inertial frame on earth... Say I use an arbitrary mass unit so that 1 unit of mass will read off 1 lb. This solves the problem. I was trying to use OP's own units in an example they could relate to, not outline a technical flaw in their mass units $\endgroup$ – Ryan Franz Oct 23 '15 at 21:01
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    $\begingroup$ The notion that Ryan is getting at is known as "pounds mass" in engineering circles. To be distinguished from "pounds force". Likewise they've also been known to use "kilograms force" when talking about weight. ::shudder:: $\endgroup$ – dmckee Oct 23 '15 at 22:07

Probably convert your weight (in pounds) to mass (in kg), and convert speed in mph to speed in m/s. To convert to kg, multiply your pounds by .453592, and to convert mph to m/s, multiply your speed in mph by .44704.

Also note that people are going to try and correct your use of pounds here. Pounds are a unit of force (like Newtons), while kilograms are units of mass. However, as long as we assume we're on the surface of the earth, we can covert between pounds and mass without too much trouble. But you couldn't use the same conversion factor on the surface of the moon: You'd weigh less (fewer pounds) but still have the same mass (same number of kgs)!


protected by Qmechanic Oct 23 '15 at 20:58

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