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Homework Help: Looks like a lot of people mess up on energy

  1. Mar 16, 2006 #1
    Simple explanation here.

    Live by these words: ENERGY IS CONSERVED.

    Gravitational Potential Energy (GPE) is stored energy. It is mgh.
    That is, mass times gravity times height.

    GPE is proportional to how high the object is. (GPE~h). The higher the object is, the more gravitational potential energy it contains.

    In most cases when we deal with GPE, we will, by convention, set the lowest point to be zero GPE (to make it easier for all of us). You can set any point to be 0 GPE but the lowest point is always the best place (unless specified in a question).


    Kinetic Energy (KE) is energy in motion, motional energy. It is 1/2m(v^2).
    That is, one-half of the mass times velocity squared.

    KE is squarely proportional to how fast an object is moving.
    (KE~v^2). The faster the object is moving, the more kinetic energy it contains.

    In most cases when we deal with KE, KE is 0 when the particle is not moving. For example, you throw an object in the air. At the highest point the object reaches, we say it is not moving at the instant and thus has no kinetic energy.


    Now, we live by the words: ENERGY IS CONSERVED. By definition, the total energy of the system never changes. The potential energy and kinetic energy may change, but the total energy (KE+GPE) will not change.

    Now, when I dealt with energy, we were given a scenario that had an inital and final point.

    The forumla for the conservation of energy is:

    dKE = -dGPE (The change in kinetic energy is the negative change in potential energy)

    KE(f) - KE(i) = -[GPE(f) - GPE(i)]
    Distribute the negative sign
    KE(f) - KE(i) = GPE(i) - GPE(f)
    Separating final (f) and initial (i)
    KE(f) + GPE(f) = GPE(i) + KE(i)
    Flip it around...
    KE(i) + GPE(i) = KE(f) + GPE(f)

    And there we have it folks, the conservation of energy in simple terms. Now let's talk about an example problem using conservation of energy.


    Let's say I happen to throw a baseball of mass (2 kg) straight up in the air with an initial velocity (5 m/s). It reaches a maximum height (h). How would we write the equation to solve for the maximum height (h)?

    Let's look at the initial part (right when I throw the baseball). I throw the baseball with an initial velocity, thus the Kinetic Energy is NOT 0 initially. I am going to go ahead and set the point where I throw the baseball 0 GPE because: it is the lowest point, the point where we set GPE=0 is arbitrary, and because I can.

    Let's look at the final part (when the baseball reaches maximum height). The baseball is not moving at this point so Kinetic Energy IS 0. Because we've defined 0 GPE at the point where I throw the ball, GPE is defined at some height (h). Now, we can write the equation for conservation of energy.

    KE(i) + GPE(i) = KE(f) + GPE(f)

    We stated GPE(i)=0 and KE(f)=0 for reasons mentioned above. Our equation now looks like this:

    KE(i) = GPE(f)

    The initial KE is equal to the final GPE. Let's go ahead and plus the formulas in for KE and GPE.

    1/2mv^2 = mgh

    Note that in this case, the mass cancels out because the mass of the ball doesn't change.

    1/2v^2 = gh

    Solving for h gets:

    (1/2v^2)/(g) = h

    Where g is 9.8 m/s/s (the acceleration of gravity).

    If we wanted to solve for the inital velocity, we would do this:

    v = sqrt(2gh)


    To make sure you fully understand this, let's compare KE and GPE.

    As the KE of a system increases at some rate, the GPE of the same system decreases at the same rate.

    NOTE: KE doesn't equal -GPE. The change in the KE is equal to the negative change in GPE.

    To prove my point, lets look at a 'table' comparing KE, GPE, and the total energy of the system (which I will denote, W).

    http://xs72.xs.to/pics/06115/table.png [Broken]

    Note that in all cases, the total energy (W) of the system is 500 J at all times. The KE and the GPE fluctuate.

    Examine the change from Point A to Point B.

    The change in KE from Point A to Point B is (150 J - 0 J) = 150 J
    The change in GPE from Point A to Point B is (350 J - 500 J) = -150 J

    Based on the formula: dKE = -dGPE

    150 J (dKE) = -(-150 J) (dGPE)
    150 J = 150 J, thus energy is conserved.


    I hope by now you understand energy and I sure hope this topic is stickied as it is the best explanation you will ever find.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 16, 2006 #2
    Note: Elastic Potential Energy works in the same way... Instead of GPE, use EPE and the formula for EPE instead of GPE.

    You should be able to get it right :-)
  4. Mar 16, 2006 #3
    I will proceed to throw away my physics books.
  5. Mar 16, 2006 #4
    LOL. Most physics books seem to add more stuff than needed... I just explained the meat of it basically... If you know how to do this and understand it, then you can apply it to any problem of the same nature.
  6. Mar 16, 2006 #5
    Thats great, until I have a HW problem that is not of the same nature and I can't solve it because I just know how to plug and chug.
  7. Mar 17, 2006 #6
    Well, actually, think of it this way. Apply the ideas of energy.

    The total energy of the system (total work) never changes.
    The change in KE is the negative change in PE.
    KE increases at the same rate that PE decreases.
    The inital energy of the system must always equal the final energy of the system.
    Setting any point PE=0 is arbitrary, usually set it as the lowest point of system.
  8. Mar 19, 2006 #7
    *bumps this*

    Reason I bumped is more and more people are asking about problems with relation to conservation of energy, kinetic energy, gravitational potential energy, and elastic potential energy... PLEASE READ THIS! It is a VERY good explanation of how energy works and how to solve problems with it.
  9. Mar 19, 2006 #8
    That is not the formula for the conservation of energy.

    The conservation of energy simply says:

    [tex] \Delta E_{system} = E_{final}-E_{initial} [/tex]

    Your equation is only valid for your example.
  10. Mar 19, 2006 #9


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    Gold Member

    There's no need to bump threads, Da-Force.

    - Warren
  11. Mar 19, 2006 #10
    I've excluded non-conservational forces in this explanation. o_O
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