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Work-Energy vs. Kinematics

  1. Mar 15, 2007 #1
    1. The problem statement, all variables and given/known data

    I have worked out at problem using the conservation of mechanical energy and the answer is correct.

    I have also worked out the problem using kinematics and got the same answer.

    My problem is that I can't wrap my brain around why it works out using the conservation of mechanical energy.

    Here is a statement of the problem (a graphic is attached as well with my solutions and some questions):

    "A motorcyclist is trying to leap across the canyon by driving horizontally off the cliff. When it leaves the cliff, the cycle has a speed of 38.0 m/s. Ignoring air resistance what is the speed when the driver strikes the ground on the other side?"

    2. Relevant equations

    1/2mvi^2 + mghi = 1/2mvf^2 + mghf

    vf^2 = vi^2 + 2gd

    3. The attempt at a solution

    Please see attached w/ my included questions.

    Attached Files:

  2. jcsd
  3. Mar 15, 2007 #2


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    Gold Member

    If I am reading you question correctly, you are wondering why conservation of mechanical energy works, since gravity does work. Is that your question?

    If so, conservation of mechanical energy works because the whole concept of Gravitational Potential Energy is, in itself, a way to express the work done by gravity. You do not need to consider the work done by gravity because it is already accounted for by the potential energy terms that are in the equation.

    Does this help?

  4. Mar 15, 2007 #3

    Doc Al

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    Staff: Mentor

    To answer your question, to find the work done by any force (gravity included) you need only count the component of force in the direction of the displacement (or, equivalently, the component of displacement in the direction of force). This can be expressed mathematically as:

    [tex]dW = \vec{F}\cdot d\vec{s}[/tex]

    Where s stands for displacement.

    In the case of gravity, things are easy. Since gravity acts vertically, only the vertical component of the displacement counts.

    Note that since gravity only acts vertically, the x-component of velocity is fixed. Perhaps now you can see how the kinematics solution matches the conservation of energy solution:

    [tex]V_{y,f}^2 = V_{y,i}^2 - 2g(y_f - y_i)[/tex]

    Since [itex]V_{x,f} = V_{x,i}[/itex], add it to both sides:

    [tex]V_{y,f}^2 + V_{x,f}^2 = V_{y,i}^2 + V_{x,i}^2 - 2g(y_f - y_i)[/tex]

    Which gives you conservation of energy:

    [tex]V_{f}^2 = V_{i}^2 - 2g(y_f - y_i)[/tex]

    Make sense?
  5. Mar 15, 2007 #4
    Doc Al,

    So is that to say that when I state the cons. of ME...

    1/2mvi^2 + mghi = 1/2mvf^2 + mghf

    that the vi and vf are already resultant velocities?

    How can this be proven?
  6. Mar 15, 2007 #5


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    work and energy are scalar quantities; they have no direction associated with them; they are positive or negative numbers (or 0). KE is always positive because of the square term of 'v", which is not a vector, but rather, a speed, the magnitude of the velocity. If a particle of mass m is moving at 10m/s, it's kinetic energy is the same whether it is moving up, down ,sideways , or any way.
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