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Work-Energy Theorem

  1. Feb 19, 2008 #1
    [SOLVED] Work-Energy Theorem

    I'm sorry that I do seem to require quite a bit of help, but sometimes when the formulas just aren't working for me I get desperate.

    1. The problem statement, all variables and given/known data

    A force is applied to a 2.5kg- radio-controlled model car parallel to the x-axis as it moves along a straight track. The -component of the force varies with the x-coordinate of the car as shown in the figure .Suppose the model car is initially at rest at x=0 and [tex]\vec{F}[/tex] is the net force acting on it.

    [​IMG]

    Use the work-energy theorem to find the speed of the car at =3.0.

    2. Relevant equations

    W = K2-K1

    W = Fx

    K = 1/2mv^2



    3. The attempt at a solution

    (2N)(3m) = 1/2(2.5kg)v^2

    6J = 1/2(2.5kg)v^2

    12J = 2.5kgv^2

    12J/2.5kg = v^2

    [tex]\sqrt{12J/2.5kg}[/tex] = v

    v = 2.2m/s wrong

    another try

    W = [tex]\int[/tex]Fx = x^2/2 from x=0 to x = 3

    so 9/2 = 1/2(2.5kg)v^2

    [tex]\sqrt{9/2.5}[/tex] = v, v= 1.9 wrong

    I keep rearranging all the formulas with the information they've given me and still nothing, any help is greatly appreciated.
     
  2. jcsd
  3. Feb 19, 2008 #2
    In your first attempt at the solution, re-examine the definition of Work.

    Does 2N*3m accurately represent the work done?

    Remember, Work is not simply F*x rather:

    [tex]W=\int_{x_1}^{x_2} F\cdot dx[/tex]

    What does the integral represent on the curve? Check your math on the second try.
     
  4. Feb 19, 2008 #3
    units of work = Joules = Newton-meters don't they?

    and that integral would mean it's still force (2Newtons) multiplied by (x2 - 3meters) right?
     
  5. Feb 19, 2008 #4
    I think to hammer home the point he was trying to make, look at that integral, carefully

    In the case where F is not a function of x and is in the same direction as dx(like it is in this problem so don't worry about that), it's a constant and comes out of the integral, so you're just integrating 1*dx, which is x, so you get the familiar F*x

    HOWEVER, in this case you actually have a F that varies with position. Oh noes an integral you must solve? Not necessarily, remember, what IS an integral? You have a function F(x), if I integrate between two points what is that result?

    So basically 2*3 would give you the work done if the force had been 2 newtons the entire time. But it's not
     
  6. Feb 19, 2008 #5
    so it would be 2 integrals than, one at points from x = 0 to x =2 and then one from x = 2 to x = 3 correct?
     
  7. Feb 19, 2008 #6
    so the work would be 5 joules?
     
  8. Feb 19, 2008 #7
    Exactly

    Not so exactly. You don't actually need to work a single integral here, so long as you remember what an integral IS, really. In general if I give you an the integral from x=a to x=b of f(x), you're gonna get a number for the solution. This number is the _____ ____ the ____

    Once you find the correct work I believe the rest of the problem you worked correctly
     
  9. Feb 19, 2008 #8
    check your math again.

    It would be helpful if you wrote out exactly what you are doing to calculate the integrals. It is possible to use the figure to easily calculate these values. What can integrals be used to represent....something about area...
     
  10. Feb 19, 2008 #9
    blockwave = the number is the antiderivative of the derivative?

    integrals represent the area under a curve, the first curve being from x=0 to x=2, and then from 2 to 3.

    I'm thinking the functions are wrong, the first function from x=0 to x=2 would just be x, so the first integral would be

    F(integral from x=0 to x=2) xdx = x^2/2 from x = 0 to x =2 and it would be 2(2^2/2) = 4

    and the other integral there is no function so

    F (integral from x=2 to x=3) dx = x from x = 2 to x =3 and it would be 2(3-2) = 2(1) = 2

    2+4 = 6.

    I'm not sure how to use the force as a changing force and not a constant one...
     
  11. Feb 19, 2008 #10
    Oh soooo close

    Exactly like you did it except I have no idea why you multiplied the first integral by 2 randomly.

    It was exactly like you said, integral of x*dx from 0 to 2, which ends up being 4/2=2

    For the second one you have the integral of 2*dx, so there's no function, you're correct, the force is just a constant, 2. So it's 2*integral(dx) from 2 to 3 which you found correctly as 2

    2+2=4

    Now for useful skill #394609 you're about to learn.

    An integral is just an area under a curve, from x=0 to x=3 in that picture you have a triangle and a rectangle. The area of the triangle is just 1/2*base*height, so 2*2/2=2, and the rectangle is just 1*2=2. 2+2=4
     
  12. Feb 19, 2008 #11
    funny as I do harder and harder physics and calc I forget simple stuff like that, lol.

    thanks, I'll let you guys know if I got the answer right in masteringphysics.
     
  13. Feb 19, 2008 #12

    You have the right idea. However, your integration on the first section is incorrect.

    [tex] W_{12}=\int_0^2 xdx=\frac{1}{2}x^2\mid_{0}^{2}=\frac{4}{2}=2[/tex]
     
  14. Feb 19, 2008 #13
    beat me to it!

    Anyway, hope we have helped. The area under the curve 'trick' is very useful in introductory physics.
     
  15. Feb 20, 2008 #14
    bump to this thread, would've posted last night but connection got messed up.

    anyway, thanks a bunch on part a), but part B, which is the velocity at x=4, has me a little stumped, because I'm assuming there's no work because it went down to 0, but obviously it moved so it has to have some velocity.

    I tried:

    K2 = K1 + W1 = 4.05 + 4 = 8.05

    1/2mv^2 = 8.05

    mv^2 = 16.1
    v^2 = 6.44
    v = 2.5m/s but I got it wrong, anyone?
     
  16. Feb 20, 2008 #15
    You're right that no work was done to it while the force was zero

    You already found the speed at x=3 for part a, right?

    So here you have something moving along at some speed, then you...do nothing to it, and wanna know how fast it's going a meter later...
     
  17. Feb 20, 2008 #16
    I'm freakin' stupid, it's the same thing, lol.

    now part C, I'm assuming they're 2 integrals, one from x=4 to x =6 and one from x=6 to x=7; with force -1 acting on the first integral and maybe 0 force acting on the second one, or could I just use the triangle formula again with 1/2*base*height = 1/2(-1)2 = -1?
     
  18. Feb 20, 2008 #17
    problem is solevd, used integral from x=4 to x=6 with the function 1/x so got ln6-ln4 =
    .405N as the force and used newton's law to find the acceleration of the particle:

    a = F/m = .405N/2.5kg = .162m/s^2

    then used v2^2 = v1^2 + 2(a)(d) to get:

    1.8^2 + 2(.162)(3) = 2.3, v = (2.3)^(1/2) = 1.5m/s

    thanks for all the help!
     
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