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Homework Help: Work and energy?

  1. Apr 6, 2012 #1
    not really. specific homework questions, more like, needing help to get started.

    There are a few things I'm having difficulty understanding..

    1)how to determine what is Ei and what is Ef.
    i.e. if a ball is shot into a net with an initial speed of 5m/s at a height of 2m, what speed will the ball travel into the net at a height of 3m?

    from what i learnt: Eti=Etf
    Eki + Eg = Ekf + Egf ..
    I don't get why though..

    cause then theres stuff like; a mass is placed on level drictionless surface, and is released of a spring attached to the side of the wall. so the energy here would be Eei=Ek+Eef.. so theres no gravity here? and when the spring is compressed towards the wall, its still considered elastic energy?

    2) a 50 kg crate is pulled 40 m along a horizontal floor by a constant applied force of 100 N, the floorr exerts a force of 50N on the crate. how much work is done on the crate by the floor. the equation we used was W=Fcos#*d ... but why would we sub in 40 m into d? the frictional force didn't push the mass to 40 m.. if more, it'd be restricting it from travelling it more than 40, so it shud be pulling back a certain distance to keep it at 40, not making it 40m itself..
  2. jcsd
  3. Apr 6, 2012 #2


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    hi wow22! :smile:

    (have a mu: µ and try using the X2 button just above the Reply box :wink:)
    in your closing example, yes there's no Eg in the formula, but that's only because it would be the same on both sides

    so no point in mentioning it! :smile:

    (and yes it's called elastic energy whether the spring is compressed or stretched)

    generally, we use all the energy that changes from i to f, and we leave out all the energy that doesn't

    the important thing is that Etotali = Etotalf

    just use common-sense to tell you what to include :wink:
    you were told wrong, the equation should be W = minus Fcosµ*d …

    work done = force "dot" displacement, the displacement is always opposite to the friction force, so the work done is always negative (the mechanical energy decreases)
  4. Apr 6, 2012 #3


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    yes, this is the Conservation of Energy theorem, which states more or less that energy cannot be created or destroyed, just converted to different forms of energy. So whatever energy you start with, you end with, since total energy cannot change.
    as long as there are no non conservative forces acting that do work (a non conservative force is for most part any force except gravity or spring force), then the total energy of a system is just the sum of it's kinetic and potential energies.
    well, there is no change in gravitational potential energy because of the level surface...you can put it in as 0 initial and 0 final if you want..
    yes, it has initial elastic potential energy when it is compressed
    Since the force of friction acts continuously through the 40 meters, that's the number to use...work will be negative because the force act opposite the displacement (theta = 180 degrees)...
  5. Apr 6, 2012 #4
    Kay thanks for the replies. I think i get that.
    I don't really want to write up a new thread so i'll just ask another question here. it's still energy..

    a pen of mass 0.057kg slides across a horizontal desk. In sliding 25 cm its speed decreases to 5.7 cm/s. the force of kinetci friction exerted on the pen by the desk has a magnitude of 0.15 N. find initial speed.

    I don't get why thermal energy equals delta kinetic energy..
    Ek2-Ek1 is work.. and work= Thermal energy?
  6. Apr 6, 2012 #5


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    work done = change in mechanical energy

    mechanical energy is kinetic energy + all forms of potential energy (gravitational, elastic, etc)

    since total energy is constant the loss in mechanical energy must equal the gain in non-mechanical energy … and that's mostly thermal energy :wink:

    (so work done also = change in thermal energy)
  7. Apr 6, 2012 #6
    Wow22, think of friction as energy "leaking out" of a system in the form of heat.
  8. Apr 7, 2012 #7
    Uhmmm I think i get it... :s Thanksss!

    so i started doing the spring questions..
    and what kind of threw me off..was the formula itself.
    we learnt that k=F/x .. so when they asked me to find the net force, i assumed that it would be, F=kx.. but its supposed to be.. Efx=F-kx? O_O" i dont see how that equation derived from k=F/x ..

    thanks in advance,
    & thanks to the previous answerers again (:
  9. Apr 7, 2012 #8


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    sorry, i'm confused :redface:

    what question are we talking about? :confused:
  10. Apr 7, 2012 #9
    OOh sorry, its a new question.

    suppose the question was; a student stretches a spring horizontally a distance of 15 mm by applying a force of 0.18 N [E], what would the force exterd by the spring be on the student; then you'd use Fx=k*x.

    But, when the question asks to find the net force of i.e. a ball of mass 0.075 is hung from a vertical spring that is allowed to stretch slowly from its unstretched equilibrium position until it comes to a new equilibrium position 0.15m below the inital one. find the net force on the ball when it has dropped 0.071

    ..the book uses Efx= mg+(-kx)

    is that a completely different equation than the Fx=kx?
    cause we never learn the Efx one in class.

    thanks !
  11. Apr 7, 2012 #10


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    but the net force on the student, or on the thing at the end of the spring that he is holding, is zero
    (why Efx ? E is energy, isn't it? don't you mean Fx ?)

    yes, if the ball was in the equilibrium position, the vertical net force would be mg - kx = 0

    the vertical net force is always mg - kx if the spring is vertical :confused:
  12. Apr 7, 2012 #11
    Idk.. the book says Efx :s like net force..energy? net energy? but that's what the textbook put. so if it was in equilibrium..thered be net force.
    but theres no net force when you're applying force? :s

    wait.. theres a difference between verticle net force and horizontal? so if vertical is mg - kx. then what's horizontal? :S
  13. Apr 7, 2012 #12


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    well, horizontal would have no mg,

    so if the spring was horizontal there would only be kx, plus any applied forces (eg someone pulling)
  14. Apr 9, 2012 #13
    Thanks for all the help tiny-tim!
  15. Apr 9, 2012 #14
    and the questions have yet to end ..

    a bungee jumper of mass 64.5 kg is standing on a platform 48 m above a river. the length of the unstretched bungee cord is 10.1 m . the force constant of the cord is 65.5 N/m. the jumper falls from rest and just touches the water at a seed of zero. The cord acts like an ideal sring. Use conservation of energy to determine the jumper's speed at a height of 12.5 m above the water on the first fall.

    So what i thought.. was that it was Eg+Ek=Eg2+Ek2
    and I solve for v2 of EK2.
    so: sqrt ( 2(g(h1-h2)+0.5v^2) and ended up with like. 26m/s or something like that. but the answers 6.37 m/s? O_O"

    I think we're supposed to use the 10.1 m of cord somewhere in there?
  16. Apr 9, 2012 #15


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    no, you need all the mechanical energy!! :rolleyes:

    Eg+Ek+Espring = Eg2+Ek2+Espring2 :wink:

    try again :smile:
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