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Force, work, and energy problems

  1. Jul 20, 2004 #1
    Force, work, and energy problems!!

    :mad: k im kind of pissed off because i just started getting the hang of this physics stuff and understanding everything going on with momentum, acceleration and velocity changes, etc...

    but now that we moved on to forces, energy, and work, its been nothing but headaches for me :frown: .

    for example, 1 question asks me....
    A force acts on a 2.0 kg cart while the cart moves 1.2 m. the work done by the force during the motion is 3.0 J. When the force begins to act on the cart, the cart is already moving with a speed of 1.0 m/s.

    (a) What is the kinetic energy of the cart at the end of the 1.2 m motion? show calculations, etc...

    what i did for this was say E<i> was KE<i> and E<f> was KE<f> so i said KE<i> = KE<f> which makes no sense because if there is an extra force they cant be equal... :cry: can anyone help?!

    (b) What is the speed of the cart at the end of the motion? Show calculations, etc...

    (I could not answer this question because i dont know how to do part A)

    SOMEONE HELP ME!!!!
    THANKS! :redface:
     
  2. jcsd
  3. Jul 20, 2004 #2
    Energy is not conserved because a non-conservative force (the push force) acts on it. So [tex]KE_i \neq KE_f[/tex]

    KNOW YOUR WORK-ENERGY THEOREMS. In non-calcululs langauge (that is, for constant forces):

    [tex] W_{\rm NC} = F_{\rm NC} d \cos(\theta) = \Delta E[/tex],

    which says that the work done by a nonconservative force changes the total mechanical energy of an object by [tex]\Delta E[/tex].

    For conservative forces,

    [tex] W_{\rm C} = F_{\rm C} d \cos(\theta) = -\Delta ({\rm PE})[/tex]

    which says that the work done by a conservative force changes the total potential energy of an object by [tex]-\Delta ({\rm PE})[/tex].

    So use the first work-energy theorem to find [tex]\Delta E[/tex]. Since common sense says that

    [tex] \Delta PE + \Delta KE = \Delta E[/tex]

    and [tex]\Delta PE = 0[/tex], then you should be able to find the change in kinetic energy of the object.
     
    Last edited: Jul 20, 2004
  4. Jul 20, 2004 #3
    it looks like all the information u posted is factoral and important....but unfortunately im only in physics 1, lol...and we're not even getting close to what equations n stuff u just told me....u r telling me about change in energy and the question doesnt ask that, asks for energy at end motion or whatever....

    the equations the professor gave in class were...

    Work = force x displacement
    K<i> + P<i> = K<f> + P<f>
    KE = 1/2 mv^2
    and finally PE = -mgh

    with these equations ro something close to it, can someone help me out?

    THANKS
     
  5. Jul 21, 2004 #4
    If you know the initial energy, then if you know the change in energy you will know the final energy. For any property (which we will call [tex]P[/tex]),

    [tex] P_i + \Delta P = P_f[/tex]

    NO, only if there is no work being done by non-conservative forces. And in the example you provided, there is work being done by a non-conservative force.

    Equations must be considered within their context. You simply cannot use the conservation-of-energy equation in the problem you provided because total mechanical energy is not conserved.

    So the initial energy is not the same as the final energy. In fact, the final energy is greater than the initial energy. By how much? According to the work-energy theorem I wrote, by the amount of work done by the non-conservative force.

    So fine this work by multiplying the force times the displacement. Add your result to the initial energy and you will obtain the final energy.

    (BTW, your instructor is assuming that the force and displacement act along the same line, thus eliminating the cosine term in the definition of work.)
     
  6. Jul 21, 2004 #5
    First list out what u know, it always helps me when I do.

    [tex]m=2kg[/tex]
    [tex]d=1.2m[/tex]
    [tex]W=3J[/tex]
    [tex]V_0=1m/s[/tex]

    Now figure out how fast the cart accelerates
    [tex]F=ma, W=Fd=mad[/tex]
    [tex]3J=(2kg)(a)(1.2m)[/tex]
    [tex]a=1.25m/s^2[/tex]

    Now use a constant acceleration equation to find the speed at the end.
    [tex]V^2 = V_0^2 + 2ad[/tex]
    [tex]V^2 = (1m/s)^2 + (2)(1.25m/s^2)(1.2m)[/tex]
    [tex]V = 2m/s[/tex]

    Now figure out its KE at that speed.
    [tex]KE = (1/2)mv^2 = (1/2)(2kg)(2m/s)^2[/tex]
    [tex]KE = 4J[/tex]
     
    Last edited: Jul 21, 2004
  7. Jul 21, 2004 #6
    Armo, the problem appears to have been set up for a work-energy solution. And besides, the work-energy solution is much, much easier than using acceleration.
     
  8. Jul 21, 2004 #7
    Well, i dont know, as long as i got the answer right
     
  9. Jul 21, 2004 #8
    AFAICT, your solution is fine. But I think the teacher wants the student to be able to use the work-energy theorems. So using acceleration would defeat the purpose.
     
  10. Jul 25, 2004 #9
    Hey 'Physicshelpneeded', I've only just seen your thread, so this might be too late for your assignment, but anyway ...

    Did your teacher talk about the Work-Energy Theorem? It says: The total work done is equal to the change in kinetic energy. So, [itex]Work = KE_{final} - KE_{initial}[/itex].

    If you haven't come across this yet, you have to go the hard way, which is what 'ArmoSkater87' did. The fact that the question gave you a distance almost tells you that this is probably the case, because you wouldn't actually need this information for the W-E Theorem.


    Just supposing he did mention the Work-Energy Theorem, it's much simpler. The only other equation you need is the definition of kinetic energy: [itex] KE = (1/2)mv^{2}[/itex].


    Part (a) asks for the final kinetic energy. Now, we are given the work, the mass and the initial speed, so writing out the W-E thm in terms of known variables (and using the defn of KE), we get [itex]Work = KE_{final} - (1/2)mv_{initial}^{2}[/itex]. A little re-arranging and plugging-in of values will give you the answer.

    Part (b) then asks you for the final speed. Since you now know, from part (a), the value of the final KE, you can use the defn of KE again to get the final speed, since [itex] KE_{final} = (1/2)mv_{final}^{2}[/itex].

    Notice that, we didn't need to know anything about the nature of the force acting on the cart at all. Was it conservative or not? Doesn't matter! Was it gravity, or friction or something combination of forces? Doesn't matter!
    All we were given the work done, the mass and an initial speed. This is the beauty of the Work-Energy Theorem.

    Hope that helps!
     
  11. Jul 26, 2004 #10
    Work = Force x Distance.
    3 = Force x 1.2
    3/1.2 = Force.
    Force = 2.5N

    As we are have no detail of any coefficient of friction or incline we are to assume energy is conserved.

    Thus, KE at start = .5(2x1) = 1J
    KE at end = 1J + 3J = 4J. (1J originally and an extra 3J inputted. Energy has not been spent on work against friction or an incline).

    KE at end = 4J


    4J KE at end.
    KE = 0.5mv^2
    4 = .5x2xV^2
    4=v^2
    V=2m/s
     
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