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Work, Energy, and Acceleration on a Frictionless Plane

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a block of mass M that is pulled up an incline by a force T that is parallel to the surface of the incline. The block starts from rest and is pulled a distance x by the force T. The incline, which is frictionless, makes an angle theta with respect to the horizontal.

    1.1 Write down the work done by the force T.


    1.2 Calculate the potential energy of the block as a function of the position x. From the work and the potential energy calculate the kinetic energy of the block as a function of position x.


    1.3 Calculate the acceleration, a, of the block as a function of position x. Calculate the velocity of the block from the acceleration and hence the kinetic energy as a function of position. Show that the kinetic energies calculated in these two ways are equivalent.


    1.4 Calculate the ratio of the potential to kinetic energy. What happens to this ratio for large T? Check your answer for the case theta = pi/2 which you should be able to recalculate easily.



    2. Relevant equations

    W=Fd
    W_net=deltaK
    U_grav=mgh
    K=(1/2)mv^2

    I think that mostly covers the relevant equations.

    3. The attempt at a solution

    For 1.1, I figured that the work would equal the force, T, times the distance over which it acts (parallel to the plane), which I called x. Hence, I found:

    W=T*x.


    For 1.2, the potential energy U=mgh would need to be adjusted for the plane, where h would be x*sin(theta). Thus:

    U=m*g*x*sin(theta).

    Now here it gets fuzzy for me. If W=deltaK, and K_initial=0, wouldn't kinetic energy K=T*x?


    So then I tried forgoing the last part of 1.2 in favor of starting 1.3 to see if I could gain some insight into what my answer for 1.2 should be. I decided that the acceleration of the block should come out to:

    a=T/M-g*sin(theta), which gives us a constant that is not dependent on position x.

    I then switched the acceleration over to dv/dx=T/M-g*sin(theta) and integrated to get:

    v(x)=(T*x)-g*x*sin(theta)

    I tried to plug this function v(x) into the equation for K=(1/2)*m*v^2, but I ended up with a really messy equation that didn't seem to get me anywhere.

    Obviously, I've screwed up somewhere in this problem, but I'm not sure where, so I don't know how to fix it. I've got a lot on my plate right now in terms of assignments due before break and a big chem exam tomorrow night, so any help is very much appreciated. Thanks.
     
  2. jcsd
  3. Mar 18, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I don't think you have the kinetic energy right. The kinetic plus the potential = Tx so K = Tx - mgx*sin(theta).

    For velocity, you let a = dv/dx but it should be a = dv/dt, which is awkward.
    I used v^2 = 2ad (good for constant acceleration) and got an expression that works - it makes the K = 1/2*m*v^2 expression equal to the Tx - mgx*sin(theta).
     
  4. Mar 18, 2009 #3
    Alright, that makes sense. Oddly enough, I actually solved for the kinetic energy correctly the first time, but I couldn't get an acceleration expression to match so I assumed it was wrong and changed it to just the T*x; I let the W=deltaK confuse me.

    Thanks for clearing things up!
     
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