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Simple Work/Energy Problem

  1. May 15, 2008 #1

    danago

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    Gold Member

    [​IMG]

    Ive solved it and the magnitude of my answer was correct, but the sign was incorrect. Ill show my working using letters rather than actual values.

    Since energy is added the the object by the force F which does work on the object, the net change in energy of the object will be Fx. The only energies of the object which are changing throughout the motion are its gavitational potential energy and its kinetic energy, so the following statement must hold true:

    [tex]
    Fx = \Delta T + \Delta V_g
    [/tex]

    Where T is the kinetic energy and Vg is the gravitational potential energy.

    [tex]
    Fx = 0.5m\Delta (v^2 ) + mg\Delta h \Rightarrow \Delta h = \frac{{Fx - 0.5m\Delta (v^2 )}}{{mg}}
    [/tex]

    Substituting all the given values into the solution derived above gives me an answer of ~0.09, but the solutions say it should be -0.09.

    Where am i going wrong?

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. May 15, 2008 #2

    danago

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    My solution seems to make physical sense as well. Since the net change in energy due to the external force is greater than the change in kinetic energy, the potential energy must also increase which corrosponds to an increase in height.
     
  4. May 16, 2008 #3

    danago

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    The solution derives the exact same formula, but somehow is supposed to give a negative answer rather than positive.
     
  5. May 16, 2008 #4
    I think your solution is correct. Sometimes the book is wrong.
     
  6. Dec 2, 2009 #5
    How can that be true if the applied force is horizontal? How is that force affecting the gravitational potential energy?

    Work = [tex]\Delta[/tex]KE = [tex]\int[/tex][tex]\vec{F}[/tex] [tex]\bullet[/tex] d[tex]\vec{r}[/tex] = [tex]\int[/tex]Fdrcos(0) = [tex]\int^{v_{f}}_{v_{i}}[/tex]vdv = (1/2)m([tex]v_{f}^{2}[/tex] - [tex]v_{f}^{2}[/tex])

    From the picture, it looks like the initial and final velocities are at non-horizontal angles, so wouldn't you have to find the horizontal components of those velocities to use KE?

    Also, assuming state 1 is the initial position and state 2 is the final position, from the picture it looks like the change in height should be negative, since the object moves downhill.
     
  7. Dec 3, 2009 #6
    I solved the problem and obtained the answer of -0.09m. To do this is assumed that the velocity at state 1 is in the x direction and the velocity at state 2 is in a random direction. I then set up my work kinetic energy equations for both the x and y directions. In the x i got Fx = 1/2(3.2kg)(vf)^2-1/2(3.2kg)(8.8)^2= (132N)(.8m). In the y direction i got (9.8m/s^2)*h = 1/2(Vf)^2. The final equation to tie it all togther is just ((Vfx)^2+(Vfy)^2)^.5 = 11.9m/s . This last equation is just setting the known final velocity magnitude of 11.9m/s equal to the square root of the sum of the squares of its components. Thus from the first two equations we can find (vfx)^2 which is just a number and (vfy)^2 which is a function of height. Hence the only unkown is the h and it will turn out negative.
     
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