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Finding final speed of object on incline

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data
    An object of mass m is allowed to slide down a frictionless ramp of angle Θ, and its speed at the bottom is recorded as v. If this same process was followed on a planet with twice the gravitational acceleration as Earth, what would be its final speed?

    Multiple Choice Options:
    (A) 2v
    (B) v√2
    (C) v
    (D) v/ √2
    (E) v/2

    I know that the right answer is A, but I do not understand why... I will explain what I attempted to do in my attempt at solution.

    2. Relevant equations
    Newton's second law: F = ma, with the force of gravity parallel to the incline being Fg*sin(Θ).
    This means that my acceleration will be g*sin(Θ)

    3. The attempt at a solution

    So let's say that we have a distance Δx, which represents the length of the ramp at the angle.
    I attempt to use a kinematic equation to solve for the length of the ramp and then try to apply the calculated result in another kinematic equation using the doubled value of acceleration.

    My attempt is attached in a photograph, and I got B. I do not understand why what I did is incorrect and why A is correct.

    The solutions manual (which I am attaching the solutions to here) says something different, and while I understand their approach.. I believe that they are basing it on the faulty assumption that with the doubled acceleration, the amount of time that the object will take to reach the base of the ramp is the same.

    Could anyone please explain why my approach is wrong and why, and what I could possibly do instead?

    Thanks in advance

    Attached Files:

  2. jcsd
  3. Jan 13, 2016 #2


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    Staff: Mentor

    I won't try to decipher your sideways photos (we prefer to see work typed in so that helpers can quote and comment on individual lines), but I agree with your conclusion that the answer should be B.

    The standard equation that they should have referenced and used here is ##v_f^2 - v_i^2 = 2 a d##.
  4. Jan 13, 2016 #3
    I do not know how to use the equation formatting tools on this site that well. Sorry.

    However, why are you of that opinion?
  5. Jan 13, 2016 #4


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    Staff: Mentor

    The LaTeX equation formatting syntax takes a bit of getting used to, but it's not all that hard for basic stuff. You can also use the symbols and special characters available via the ##\Sigma## icon and use the ##x_2## and ##x^2## buttons to achieve subscripts and superscripts.
    Which opinion is that?
  6. Jan 13, 2016 #5
    The opinion that the answer should be B?
  7. Jan 13, 2016 #6


    User Avatar

    Staff: Mentor

    Because I look at the equation I quoted and can "see" the answer: The standard equation that relates velocity to acceleration and distance involves velocity squared being proportional to the acceleration multiplied by the distance, or turning that around: for a given fixed distance the velocity is proportional to the square root of the acceleration.
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