The Work-Energy Theorem -Or- “Jacob starts writing about actual Science, again…”

Welcome back to the grind, PaleoPosse!  Did you have a good Thanksgiving vacation?  I sure hope so, because today I’m going to flex your brain muscle and attempt to teach you some REAL science: PHYSICS.  Not that mushy, gushy “biology” and “paleontology” pseudo-science that Ryan and Patrick peddle.

Microbes ain't got SHIT on me

AND if you stick through it all the way to the end, there’s a prize in store for somebody!!!

But enough of that, today I’m going to teach you about the Work-Energy Theorem.  Otherwise known as, “Everything they tried to teach you in high school physics.”

Work & Energy – What are they?

In order to understand the Work-Energy theorem, you need to understand the separate definitions of both Work and Energy.

Work = Force x Distance

Taken in context, Work is an action that is performed on an object or system.  So a continuous force applied to an object over a given distance constitutes “Work” performed on that object.

To illustrate, I will call upon my friends Dink and Donk.  Dink and Donk don’t really care about physics, but they like to have fun.  So D&D decide to go do some Jackass-style shopping cart races (or shopping carriage races, if you’re from the NorthEast) at the local supermarket.  Dink is in the shopping cart and Donk pushing him.

PowerPoint is my Photoshop

Let’s assume that Donk can push Dink and the cart with 10 lbs of force. He pushes for 40 feet, because that’s how much parking lot there is until the cart hits the curb.

EXTREEEEME!!!

If Donk maintains 10 lbs of force over the entire 40 feet, he will have performed 400 ft-lbs of Work.

10 lbs x 40 feet = 400 ft-lbs of Work
(or 542.3 Joules for those of us who like to work in SI units)

So when Dink and Donk get home and their father berates them for jerkin around, and tells them to get a job, Donk can legitimately say, “But Dad! I did a lot of Work out there! 400 ft-lbs of it!”

So now that you understand Work, what is Energy?

Energy = 1/2 x Mass x Velocity²

*Technically this is just “Kinetic” energy.  In reality, Energy can be expressed in kinetic energy, potential energy, thermal energy, pressure changes, or chemical energy.  But for the purposes of today’s lecture, we’re going to focus on just Kinetic energy.

So taken in this context, Kinetic Energy is a characteristic which an object or system has at a given point in time(as opposed to Work, which is an action performed on an object or system). So if you know an object’s mass and it’s velocity, then you can accurately calculate that object’s current Kinetic Energy.

In the case of Dink and Donk, we can can determine their Kinetic Energy at any point along their 40 foot journey.

Force vectors are Red, Velocity vectors are Purple. FACT.

For the sake of simplicity, let’s say Dink and Donk both weigh 150 lbs (4.662 slugs, the English unit for Mass), and the shopping cart weighs 20 lbs (0.622 slugs).  Let’s also assume that Donk is a pretty fast runner, and can get this puppy up to around 7 miles per hour (or around 10 feet per second).  So what is their combined Kinetic Energy?

1/2 x (9.946 slug) x (10 ft/s)² = 497.3 ft-lbs of Energy

(trust me, the units work out…stupid english units)

Now there’s something interesting here that I hope you’ve noticed: the units of measurement for Work and Energy are THE SAME.  Ft-lbs in our case, or Joules in SI units.  This is the essence of the Work-Energy Theorem.

The Work-Energy Theorem – Equating Work and Energy

What we have shown in the examples above is that Energy and Work are two completely different concepts, yet they are expressed in the same units.  What does this mean?

It means that Work and Energy are two sides of the same coin. Work creates Energy, and Energy performs Work.

Therefore we can equate the two formulas

Work = Energy
Force x Distance = 1/2 x Mass x Velocity²

This allows us to perform some pretty incredible calculations.

In the case of Dink and Donk, we can calculate how far Donk will need to push Dink in order to reach a speed of 10 feet per second.  All we have to do is re-arrange the formula to solve for Distance.

Distance = [1/2 x Mass x Velocity²] / Force
Distance = [1/2 x (9.946 slug) x (10 ft/s)²] / 10 lbs

Distance = 49.73 feet

Pretty cool huh?  Well shit’s about to get even cooler…

Using some incredible Math-Magic, we can transform the Work-Energy equation into a set of other equations called the “Kinematic Equations”.

Bonus points if you can show the proofs for the Kinematic Equations

Using the Kinematic equations, we can predict a plethora of different kinetic characteristics.

For instance, in the case of Dink and Donk, we can predict how far Dink will fly after hitting the curb (assuming there is a 6 foot drop after the curb, and Dink is currently 1 foot off of the ground).

For the sticklers, assume Dink does not experience air resistance. :-p

Now the following math steps are generally considered to be the most annoying calculations that a high school Physics student will ever have to do.  (Who REALLY wants to use the quadratic formula? BLEGH!)  But if you’re interested in Physics or Engineering, don’t worry! Once you add Calculus into the mix, these equations get A LOT simpler and easier to handle.  But for now, I’ll solve them for you!

So first we have to think about the problem.  We know that Dink’s initial velocity before striking the curb is 10 ft/s.  Assuming Dink does not experience air resistance, his forward velocity WILL NOT CHANGE as he falls.  In other words, his final velocity is also 10 ft/s. In addition, he is not being pushed while he is in the air, so his lateral acceleration is 0 ft/s².

Vi = 10 ft/s                     Vf = 10 ft/s                 a = 0 ft/s²

So let’s look back at our equations now… we have Initial Velocity, Final Velocity, and Acceleration – and we’re trying to solve for distance… all we need now is Time.

How long does Dink stay in the air? That is determined by gravity and the height that he falls.  Since we know that gravity accelerates downward at 32 ft/s², and that Dink has no initial vertical velocity, we can solve for the time it takes Dink to fall using the upper left equation shown in the Kinematic equations.

Understand that "d" in these calculations is the vertical height of 7 feet, not the lateral distance which we are trying to solve. The Kinematic Equations only apply to one axis at a time.

So now we have our falling time, and we can now solve for our lateral distance, d!

6.6 Feet! NEW RECORD!

Work-Energy Theorem – COMPLETE!

So look back at what we’ve done here.

  1. We calculated the Work performed by Donk while pushing Dink and the shopping cart.
  2. We calculated the Kinetic Energy of the Dink-Donk-Cart assembly.
  3. We calculated how hard Donk would have to push to get Dink and shopping cart up to 10 ft/s.
  4. We calculated how long it will take Dink to fall from the 7 foot ledge.
  5. We calculated where Dink will land after he’s ejected from the shopping cart.

Now I don’t know about you, but I’ve had entire homework assignments with less calculations than this.  This is a feat to be proud of.

But more importantly, I hope you’ve taken away a ground-level understanding of the Work-Energy Theorem.  If you know how much energy you want a system or object to have, then you can calculate how much work you need to perform to get it there. Conversely, if you know how much energy a system has, you can calculate how much work that energy can perform.  This has literally MILLIONS of real-life examples:

  • The energy of wind movement performs work when it turns a Wind Turbine
  • The chemical energy in gasoline performs work on a piston, which in turn performs work on a vehicle to create kinetic energy.
  • Work is performed on air as it enters a Jet Engine to speed up the air, which results in higher kinetic energy of the air particles, which pushes the airplane
  • Stirring a pot of water performs work in the form of heat transfer, which results in a higher temperature in the water (higher temp = higher energy)
  • When you throw a water balloon at someone’s face, their face performs work on the balloon, which then increases in pressure (higher pressure = higher energy) until it pops.

So here’s the skinny PaleoPosse.  I want to make a video demonstrating this principle.  If you can think of some really good examples of the Work-Energy theorem that I could do myself and record (without having to buy anything TOO expensive), I will pick a favorite and send them a prize!  (I do, however, reserve the right to pick NONE of your ideas, if I happen to have a really good one.

I hope you’ve enjoyed this week’s blog post! I’ll try to do these “Real Science” posts more often now.

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12 Responses to The Work-Energy Theorem -Or- “Jacob starts writing about actual Science, again…”

  1. benNo Gravatar 29 November, 2010 at 10:27 am #

    rad!

  2. Dustin K.No Gravatar 29 November, 2010 at 6:41 pm #

    There would be obvious applicability in weight-lifting terms for how much work and energy are being done/created/used during a workout, and for comparing different exercises (useless curling vs. awesome thrusters, for example…), both for total ft-lbs performed and adding in the time element for faster (Olympic) movements vs. slower (bodybuilding) techniques.

    I need to keep re-reading this post until I stop glazing over during the math… sigh. :D

    • JacobNo Gravatar 30 November, 2010 at 4:40 am #

      Yep, you could definitely show mathematically how much work is done for each work-out. You would need to measure the total displacement (which is likely to be an arc, not a straight line), and then calculate the force necessary to lift the weight through the vertical distance (i.e. fight gravity).

      Conversely, you could measure the speed of a runner or swimmer to calculate their total kinetic energy, and back out how much work they must be producing to reach that speed.

      The tricky part is in accounting for imperfections. In the case of Dink and Donk, I didn’t account for friction or air resistance. A lot of times, these sort of “nuisance factors” can be ignored, but in other cases they can’t.

      In the case of the human body, they definitely CANNOT. Our muscles aren’t optimized for energy efficiency, they’re optimized for broad-based utility. So if you calculate that a runners kinetic energy equals 400 ft-lbs, he/she may actually be performing 450 ft-lbs of work to overcome the friction in their joints, the vertical “jumping” necessary to keep moving forward, and the reciprocal muscle movements that don’t actually add to their speed. This doesn’t even begin to account for Calories burned pumping your arms, keeping your balance, and running your internal organs. (P.S. Calories are a form of energy, and ft-lbs can be converted directly to Calories… 1 Calorie = around 3000 ft-lbs)

      So for that reason, estimating Calories-burned is extremely difficult for nutritionists, because each person moves differently, and expends different amounts of energy doing similar movements.

      Regarding the math parts… Which parts in particular do you have trouble with? I’m trying to turn this into a video aimed at high schoolers, so if you could help me improve it I’d really appreciate it.

      • RyanNo Gravatar 30 November, 2010 at 4:44 pm #

        I remember when I was in high school we never did much of the math that you mentioned specifically the kinematics equations. Also to make it easier we worked in metric units, even though I live in the US. So if you are trying to gear it toward high school aged kids, maybe try something just giving them a the metric units and then show the cancellations, they never seemed very difficult to grasp.

        I think the example we did in our class was we took the teachers car out in the parking lot, with a bunch of scales and 3 of us pushed while trying to maintain the force on the scales at the same amount. We had a driver(of age) call out the speed once we crossed a point.

        After the first run, we put more kids in the car and did it again. We used the speed of the car along with the force of us pushing and the known distance of the starting line and ending point to find the mass of the car with just the driver and then the mass of the car with more people in it.

        It worked out that even tho we couldn’t really push with a consent force very well, the mass ended up being pretty close to being right. So we did the experiment and only used the Work = Energy or Force x Distance = 1/2 mass x velocity^2

        On a side note, started listning about a month ago and I can say I am obsessed with the podcast. I’m still catching up on all the episodes, but I have switched from music to Science … Sort Of for my driving to and from school over breaks and walking back and forth to classes while at college.
        I am a Physics Major, graduating in Dec and listening to you guys talk about science makes me excited to get out and go do science in the real world
        I love your guests and can’t wait to see who you get next. I hope when I get a job, I can find co-workers as fun as you guys. Thanks a bunch for doing the podcast.
        Ryan Funk, Pennsylvania

        • JacobNo Gravatar 1 December, 2010 at 11:05 am #

          That’s a great idea! Only one question: How did you verify the weight of the car? Were you able to take it to some truck-scale or something?

          Congrats on being so close to graduation!

          In hindsight, maybe the Kinematic Equations were a little too much. I just figured they were “too cool” to leave out, but I think I forgot how much everyone hates them…

          • RyanNo Gravatar 1 December, 2010 at 10:02 pm #

            All cars have a little sticker somewhere on the door, I believe the drivers side door. That gives the weight of each side, and I believe a weight of with gas and without gas. After that he just took everything out of the car and trunk and it should have been the weight that was given on the sticker. But its been a while so I forget. We just used the scales for the people in the class. A truck stop would work though. That’s actually a wonderful idea if you are going to make a video for it because you don’t need to get permission to take a class of underage children to go to a truck stop. I could see the teacher begging “But its for science!”

            I agree, on the kinematics equations. They are very cool, and once I got to college and learned calc and how to manipulate equations I adored them at every turn. They made the work easier then ever. The only problem is, unless you show your work, most people won’t understand how they work other then a “they just work” argument. So many people just get confused and lost in them until they can work through the math themselves.

  3. benNo Gravatar 30 November, 2010 at 6:36 am #

    okay, so what if there’s a weight lifter doing biceps curls. he lifts the weight up vertically. hooray. he pushes up and does work. but THEN, he uses the strength of his triceps to slowly lower the weight (and thus, work them out too).

    is he doing work on the dumb-bell? the force he’s applying is in the opposite direction from the motion.

    • JacobNo Gravatar 1 December, 2010 at 11:04 am #

      OK so you bring up 2 very important points about the concept of “Work”
      1) When there is zero displacement, there is zero Work
      2) The direction of Work and Energy always depends on your frame of reference.

      So imagine you’re holding a dumb-bell in-lace in front of your chest. You have to exert some Force in order to keep the dumb-bell there, and not falling to the ground. However the dumb-bell DOES NOT move, and therefore you have performed no Work on the dumb-bell.

      There is a tendency to equate Force with Work among some students. The difference is subtle, but important. Force is exerted BY an object, Work is performed ON an object.

      The way to figure out which is which (and figure out what frame of reference you are in), is to look at the change in energy. If there is no change in energy of the dumb-bell, then no Work has been performed on it.

      In the case of the slowly-lowering dumb-bell, it starts with X amount of gravitational potential energy (based on it’s height), and zero kinetic energy (no movement.) As you lower it, it’s gaining kinetic energy, and losing gravitational potential energy. At the same time, your arm is being pushed downward by the weight of the dumb-bell. In this case, the weight of the dumb-bell is performing work on your arm, and GRAVITY is the force creating the movement of the weight.

      So to answer your question, he is NOT doing work on the dumb-bell as it’s lowered – the dumb-bell is doing work on him.

      • LauraNo Gravatar 8 September, 2012 at 1:00 pm #

        Ok so I stumbled on this post while looking for resources for teaching the work-energy theorem (and in the process deciding this place is fun and I should follow you guys), and your response to Ben’s comment has me confused.
        By controlling the lowering of the weight, you’re limiting the amount of KE that the weight gets, compared to if you just let it drop. By affecting the energy, aren’t you doing work?

        Same with friction… an example problem in my textbook shows the force of a person pushing a box as positive work, and the friction on the box doing negative work. The explanation is that the box is moving in the positive direction, but by having a force 180 degrees from the direction of motion, you get a negative number when you calculate the work done by friction.

        Thoughts?

  4. DustinNo Gravatar 1 December, 2010 at 5:58 am #

    Oh, I just suck at math (thanks to a teacher in 11th grade who made pre-calc/trig WAY too hard, tanking my GPA, but allowing me to get a 99 on the state Regents exam… I quit math in protest, and have been kicking myself ever since…)

    The workout program I follow (http://www.crossfit.com) talks a lot about intensity. One of their hallmarks is certain workouts that never change, and the only goal is to complete them FASTER than the last time you did it. Therefore, total work goes up as force remains constant and time shrinks.

    Within reason, especially for those exercises that use barbells/dumbbells/kettlebells, the math should be somewhat easily applicable to the weight itself, leaving out the effects you listed in the athlete themselves.

    For example (and feel free to correct any obvious math errors), doing 45 deadlifts with a 102kg barbell results in the barbell moving approximately .6m straight up (done correctly). That results in a workload of 2,754m-kg. Do that in 5 minutes, and you are doing 550.8m-kg/minute. Do it in 2 minutes, and you’re doing 1,377m-kg/minute, which is obviously a much higher power output.

    My question is, how does that then reflect the force equations? Can you use the distance and number of reps during a period of time to calculate the acceleration needed for the force equations? Or would that have to be measured empirically somehow? (I’m way over my head at this point…so feel free to point out if the the correct answer is akin to Feynman talking about why you can’t actually describe magnetism…) :D

    • JacobNo Gravatar 1 December, 2010 at 11:31 am #

      You’re giving me really good feedback, Dustin, I really appreciate it.

      OK so let’s address a few points here:
      1) Power vs. Work
      When you talk about doing the same exercise faster, you’re talking about increase the POWER of your muscles. The equation for Power is simply Work/Time (P = W/t). So the faster you do the same amount of work, the more Power you have. And the more Power your muscles have, the faster they can burn Calories (i.e. do work)

      2) Work & Energy
      The version of the Work-Energy theorem which I showed in the post neglected to include gravitational potential energy. In reality, you can increase the energy an object has without changing an objects’ initial and final velocity. For example:

      Position A) A 10 lbs weight is laying on the floor. It has zero velocity.
      Position B) You slowly pick up the weight, and raise it 7ft in the air. At the end of this movement, the object still has zero velocity.

      Now, obviously the weight had some velocity between points A and B. But the total energy of the object at position B is greater than position A, because it now has greater gravitational potential energy based on it’s higher altitude. Potential energy is equal to the object’s mass (m), times the acceleration of gravity(g), times it’s altitude(h) – PE = mgh

      This is called “potential” energy, because we can use it to perform Work. A waterfall can turn a water-wheel, but a stagnant pond cant, because of the potential energy difference in the top and bottom of a waterfall.

      SO – back to your question.

      If you do the same exercise faster, you are doing the same amount of work to displace the object…HOWEVER, you have to do slightly more work to move the object faster. In the end, the difference in Work is almost negligible. The goal of your exercises is not to burn more Calories, but to increase the size and work-performing capacity of your muscles. Running 1 mile at 10 mins/mile burns around 100 Calories, and running 1 mile at 6 mins/mile burns around 100 Calories too. The difference is in the POWER of the muscles performing the work.

      And lastly, you asked “Can you use the distance and number of reps during a period of time to calculate the acceleration needed for the force equations“. Yes you can, but it gets tricky because you have to split it into a bunch of sub-component calculations. Each time your Force vector changes, you need to do a new calculation.

      (However, it’s fairly simple to calculate the Force necessary to make the weight reach a certain speed. The only difference in this calculation and the ones I’ve shown here is that you need to include the gravitational potential energy term, AND you need to add the force of gravity on the weight to your applied Force, since you have to overcome gravity first to make the object move.)

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