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Homework Help: Mechanical energy equation analysis

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Explain what is meant by the conservation of mechanical energy. Write the corresponding equation in detail (in terms of mass speed ect.)

    a) If friction is not truly negligilble, how would you insert the influence of friction in the equation of the above question? Write the corresponding equation in detail.

    b) If friction is not negligible which should be the larger megnitude, potential energy or kinetic energy?

    2. Relevant equations
    KE= .5(m)(v)^2
    E=PE+KE (I am not sure if this is the right equation for mechanical energy?)

    3. The attempt at a solution

    I need help with this part. I am not sure what they mean by the conservatiob of mechanical energy equation, do they want KE=-PE or E=KE+PE or some other equation. Also I am unsure about how to addd friction into the meechanical energy equation, basically I need help with everything!!! Please help!!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 8, 2008 #2
    They are asking you to explain what part friction plays in the conservation of energy.

    For instance, you know that energy is always conserved, ie. energy changes between potential and kinetic (and many other types) but you can never create or destroy it.
    Friction affects energy within a system by doing work.
    I hope this gives you a hint.
  4. Oct 8, 2008 #3


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    E is the total mechanical energy of the system and is a conserved quantity.

    [tex]\Rightarrow E_{initial}=E_{final}[/tex]

    Start by using the relationship E=PE+KE to write this conservation equation in terms of [itex]PE_{initial},PE_{final},KE_{initial}[/itex] and [itex]KE_{final}[/itex]
  5. Oct 8, 2008 #4
    alright, so I have the conservation of mechanical energy to be KEi+PEi=KEf+PEf. How do I then Incorporate friction into the equation?
  6. Oct 8, 2008 #5
    well perhaps KEi + PEi = KEf +PEf + Friction
    When you think about it, potential will basically never be entirely converted to kinetic or vice versa, there will always be energy converted into other types, commonly frictional heat (thermal energy).
  7. Oct 8, 2008 #6
    okay I get it, the loss in potential and kinetic energy is then gained by the new thermal energy, friction.

    So how do I break down friction to include mass velocity ect?

    I know that the equation can look like this:

    .5(m)(vi)^2+mg(yi)=.5(m)(vf)^2 +mg(yf) +friction

    What is friction broken down into?
  8. Oct 8, 2008 #7
    It can be looked at in a few different ways but the best way is like this. Work is [tex] F\Delta d [/tex]

    Friction is generally measured by the coefficient of friction ([tex] \mu [/tex]) multiplied by the normal reaction force. The coefficient has no units therefore the product is still measured in Newton.

    Kinetic friction occurs over a distance, considering the particle being affected is moving. Therefore you can say that the work done by friction is equal to [tex] \mu F\Delta d [/tex]
    Where F is the normal reaction force of said particle.

    However if you're talking about static friction it becomes slightly more difficult because the particle is not moving when it's being affected by this form of friction.
  9. Oct 8, 2008 #8
    I see!! Thank you so much!!! I have been trying to wrap my head around this problem for a while.

    So if friction is not negligible what is bigger, potential energy or kinetic energy?

    Would my answer have to depend on the situation that would be occuring or is there a general rule?
  10. Oct 8, 2008 #9
    yes, it's completely dependent on the situation. Some scenarios have zero potential energy and all kinetic whilst others have the opposite. However, nearly every scenario will be exchanges between both.
  11. Oct 8, 2008 #10
    so in my situation, where a 270 g mass is dragged up an incline with angle of .55 degrees by a falling mass of 10 g, which would be bigger?
  12. Oct 8, 2008 #11
    Well I'm going to have to assume there is no friction (because you did not give a coefficient) and I think possibly it's 55 degrees not .55 degrees but I could be wrong.

    Are we talking about potential and kinetic energy of the system or of each individual mass?

    Mass 1, the mass pulling down = 0.01 kg
    It is pulling straight down, it's only force components are the y axis. Therefore it's pulling with force mg.

    Mass 2, on the incline = 0.27 kg. It has an x and y component. The component pulling it back down the incline will be mg sin(55) and the component that is the normal force will be mg cos(55).

    Now you can work out the net force, which will be +m1g - m2g sin(55) = net force.

    They are subtracted because they are in opposite directions.

    With the net force you can find the net acceleration. You know that the system was initially at rest so vinitial = 0

    If you knew the displacement it will undergo, the time it would take or the final velocity, we could solve the equation numerically. However without this information we cannot.

    But we can say that because it started off at rest, kinetic energy = 0 and gravitational potential was at its maximum. As the masses started to move, m1 lost gravitational potential (got closer to earth) and gained kinetic, while m2 gained a little potential energy (less than m1 because its y component is smaller) and also gained kinetic.

    So as time goes on, energy is transformed from potential to kinetic until eventually (provided m2 falls off the incline and hits the earth) all energy is kinetic and none is potential (the moment before it hits the earth).
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