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Calculating final velocity of an object on an inclined plane

  1. Oct 26, 2011 #1
    Hello everyone, I'm stuck on a dynamics review question. I was told to solve like I would with other inclined planes, however mass was not given. I am not sure how to proceed.

    1. The problem statement, all variables and given/known data
    A roller coaster reaches the top of the steepest hill with a speed of 1.4 m/s. It then descends down the hill, which is at an average angle of 45° and is 50 m long. What will its speed be when it reaches the bottom? (Answer: 26 m/s)

    2. Relevant equations
    Fnet=ma
    Fg=mg

    3. The attempt at a solution
    I drew a free body diagram, separating the x and y components of Fg. This is futile as mass is not given, thus Fg, Fn, and Fnet cannot be calculated. Am I missing something here?
     
  2. jcsd
  3. Oct 26, 2011 #2
    Hello, if you are familiar with conservation of energy, here is a method for solving this problem.
    [itex]K_E=P_E[/itex]
    In doing so, you eliminate the dependence of mass.
     
  4. Oct 26, 2011 #3
    Could you elaborate? Thanks. :)
     
  5. Oct 26, 2011 #4
    [tex]\frac{1}{2}mv^2=mgh[/tex]

    Which in words states that the kinetic energy at the bottom is equal to the change in potential energy.

    The height, h, can easily be determined as you know the angle of incline.
     
  6. Oct 26, 2011 #5
    So, in that case I would assume mass is negligible and leave it out of the equation?

    Here we go:
    First I manipulated the formula given (assuming mass in negligible) to get √2gh=v
    50cos(45°) = 35.4 m = height of the coaster to the ground
    √2(9.81m/s2)(35.4m) = v
    v = 26.3 m/s rounded off to 2 sig digs is 26 m/s
    Thanks bro! You've been a big help. :)
     
  7. Oct 26, 2011 #6
    The mass is not negligible. I would assume the mass of a rollercoaster is quite large relative to you, or I. However, the change in energy for this particular instance is not dependent on the mass of the system. Hence, the m is cancelled upon manipulation of the equations.
     
  8. Oct 26, 2011 #7
    Right, because dividing m by m yields 1. This will be of great aid on my unit exam!
     
  9. Oct 26, 2011 #8
    I wouldn't advise applying any theorms that you are not familiar with. You can also apply kinematics with this question to solve for the velocity. For example, one can apply
    [itex]v_f^2-v_i^2=2a_y\Delta x[/itex]

    The acceleration is due to gravity. The intial velocity in the y-drection is 0. The change in position is the same as you had used before.
     
  10. Oct 26, 2011 #9
    In that case then, to solve for vf I would manipulate to get
    vf = √vi2+2ad
    Which yeilds
    vf = √(1.4m/s)2+2(-9.81m/s2)(-35.4)
    vf = 26.4 m/s when rounded to 2 sig digs = 26 m/s
    I think that was exactly what I was looking for (seeing as the test is on kinematics and dynamics). I really appreciate the time you took to help me out. Thanks again.
     
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