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Help, a very basic physics problem.

  1. Aug 21, 2013 #1
    1. The problem statement, all variables and given/known data

    An object moves with the velocity of 3.7 m/s, ahead there is a "hill". How far up the "hill" will the object get before it stops? Calculate "h". Assume there is no friction.
    [Broken]


    3. The attempt at a solution

    Well, I really don't know much about how to calculate these things, I've done google searchs without results, do you know of any formulas I can use? And please if you do know the formula and decides to share it with me, explain what the certain letters mean so I don't just read, mgh=19Kpf and have no idea what anything means. I am extremely new to physics but find it very interesting (and yes I know the "equation" I wrote above isn't a real one, I was just making a point :))
    1. The problem statement, all variables and given/known data






    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2013 #2

    CAF123

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    Gold Member

    What energy does the cart possess while moving along the straight? What energy does it possess when it is on the ramp? Relate these via energy conservation.
     
  4. Aug 21, 2013 #3
    Well, since there is no friction, this implies that all the energy is conserved, with other words: energy will not get lost. So the total energy is constant.

    In this case there are two kinds of energy: kinetic and potential.

    Kinetic energy
    The kinetic energy is determined by the amount of velocity and the mass something has.
    The formula is: [itex]T=\frac{1}{2}mv^{2}[/itex]
    T=kinetic energy
    m=mass
    v=velocity

    Potential energy
    The potential energy is determined by the mass, height and the gravitational acceleration.
    [itex] V = mgh [/itex]
    V = potential energy
    m = mass
    g = gravitational accelartion (about 9.8 when we are on the earth)
    h = heigth

    Initial situation
    At the beginning we are on the ground, so our height is 0. Which makes the potential energy zero, because zero times something else remains zero. so we can say: E=T+V=T+0=T
    So we only have kinetic energy: [itex]T=\frac{1}{2}mv^2[/itex], for simplicity I will say our mass m=1. Later on I will tell you why i did that.
    So [itex]T=\frac{1}{2}*1*(3.7)^2=6.845 J[/itex] (energy is measured in Joules, or in short just "J")

    Final situation
    We get maximum height when the object stands still, just before he will fall down again. So his velocity is zero; E=T+V=0+V=V
    [itex]V=mgh[/itex]. I said that m=1, but we could also have said that it was 4756853, it doesn't matter because it would have had the same effect on the potential energy. (a little detail, ask me if you don't understand what I mean)
    So the potential energy has to be exactly 6.845 J since we stated that all the energy is conserved during the motion. From now on you should be able to calculate the height, IF you understood what I just wrote.

    Hope I didn't make any mistakes, it's pretty late here.

    Did you understand what I just wrote?
     
    Last edited: Aug 21, 2013
  5. Aug 21, 2013 #4

    CAF123

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    Gold Member

    Electric Red, this forum does not allow full solutions to questions.
     
  6. Aug 21, 2013 #5
    Oh, my mistake :blushing:. I read the FAQ but not yet all of it.
    I will edit my post, thanks!
    Could've been more embarrasing.
     
  7. Aug 21, 2013 #6
    I think so, so since V = mgh = 6.845J we can calculate h like this: m=1 * g(9.81) * h = 6.845

    That means that 9.81 * h should be 6.845. 6.845/9.81 = 0.6977m

    So the height is 0.6977m. Is this correct?

    And thank you so much for your help, I felt like you explained everything very thourough and it helped me understand it very clearly, you should become my teacher!
     
  8. Aug 21, 2013 #7
    Great! You understood what I explained, yes your answer should be something like that. You used 9.81 in your final calculation, I used 9.8. That number depends on where you are on the earth, just google "gravitational accelaration".
    Anyway, just pm me if you have more questions, or tell me when posted a new one. Glad to help you, it also helps me to rethink about stuff and I can improve my teaching-skills.
     
  9. Aug 21, 2013 #8
    Absolutely I will, thanks again! You're a great teacher. I feel like I understand it perfectly and would have no problem with any similar problems again.
     
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