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Block sliding down ramp, energy problem

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A 1.79-kg block slides down a frictionless ramp, as shown in the figure below.


    The top of the ramp is h1 = 1.54 m above the ground; the bottom of the ramp is h2 = 0.287 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance away. Calculate the distance .


    2. Relevant equations
    1/2mv^2


    3. The attempt at a solution
    I'm lost.
     
  2. jcsd
  3. Oct 13, 2008 #2
    Re: Energy

    The figure is not shown, however I can give you this advice.

    [tex] U_g = mg\Delta h [/tex]

    and [tex] E_k = \frac{1}{2}mv^2 [/tex]

    Initially kinetic energy = 0, all energy is gravitational. Let the two equations equal each other to solve for v.

    Then use constant acceleration formulae to find distance
     
  4. Oct 13, 2008 #3
  5. Oct 13, 2008 #4
  6. Oct 13, 2008 #5
    Re: Energy

    I found v easily, but I don't have an acceleration or a time to find distance traveled.
     
  7. Oct 13, 2008 #6
    Re: Energy

    Well you do, you just have to think about it.
    There is no force in the x plane once it has left the ramp, therefore velocity is constant.
    The acceleration in the y plane will be 9.81N (due to gravity)
    The distance it will fall in the y plane is 0.287m (the distance from bottom of ramp to ground).
    and the initial y velocity is zero (it specifies it is initially travelling horizontally).

    This is all the information you need.
     
  8. Oct 13, 2008 #7
    Re: Energy

    I'm sorry this isn't making sense to me for some reason, do you find time with that information given. I am trying to find the distance traveled in the x direction correct?
     
  9. Oct 13, 2008 #8
    Re: Energy

    Yes that's correct. You have velocity, and you want distance, in order to calculate this, you need time.

    You can calculate (using the parameters I defined in my previous post) the time it takes for it to fall to the ground. That will be the same time you will use to calculate your final answer.

    HINT: [tex] s = ut + \frac{1}{2}at^2 [/tex]

    where s = displacement
    u = initial velocity
    t = time
    a = acceleration
     
  10. Oct 13, 2008 #9
    Re: Energy

    Thanks for your help, but I am still getting the answer incorrect. I did all what you said


    vfx = 4.96
    t = .242
    vix = 0
    xi = 0
    xf = ?

    Right?
     
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