1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work done by a changing force

  1. Jan 9, 2014 #1
    ImageUploadedByPhysics Forums1389298689.007084.jpg


    I am stuck with the above question, my answer to a is 40J but the model answer is 30J



    Really frustrating as the book I am ready only goes over straight line graphs
     
  2. jcsd
  3. Jan 9, 2014 #2
    Remember that the work done for this type of problem is the area between the ##x##-axis and the given curve. You got that value via adding up the areas between ##x = -2## and ##x = 4##. The first part asks you to determine the total work done. You also need to consider the "negative" work done between ##x = -4## and ##x = -2##. It also takes in account for this problem.

    Try again to see if you can get ##30## J instead of ##40## J.
     
  4. Jan 9, 2014 #3
    So for the first curve can I use 1/2B*h?

    Because I done that for the two slopes and b*h for the square part?
     
  5. Jan 9, 2014 #4
    Yes, you need to use the area of the triangle formula for the region under the ##x##-axis to obtain the "negative" work done.

    Since we want to compute the total work done between ##x = -4## and ##x = 4##, we combine the works altogether to obtain the result. Don't forget this part.
     
  6. Jan 9, 2014 #5
    So I get 4 work values?
     
  7. Jan 9, 2014 #6
    So from left to right I get

    20J
    -10J
    20J
    10J

    So that gives me 40J?
     
  8. Jan 9, 2014 #7
    I didn't get those values. Let's try again.

    Remember that for the region under the ##x##-axis, regardless of the values of ##x##, you get the negative work done, using the area of the triangle:

    ##A = -\dfrac{1}{2}|b||h|##

    where the brackets indicate the absolute value of any number.

    Then, for the region above the ##x##-axis, (also regardless of the values of ##x##) you use the area of the trapezoid, which states

    ##A = \dfrac{1}{2}|h||b_1 + b_2|##

    where ##b_1## and ##b_2## are arbitrary bases of the trapezoid and ##h## is the height of the trapezoid.

    You can also compute the area of the bigger region by triangle and square area formulas as you attempted. That is: determine the area of two triangles and the area of the square.

    So in summary, you should get the following equation (ignoring the signs of ##x## values)

    ##\text{Total work done} = \text{Area between x = -4 and x = -2} + \text{Area between x = -2 and x = 4}##

    where ##\text{Area between x = -4 and x = -2}## is negative while other area is positive.

    Notes: Be careful about how you compute the first value. The shape is a triangle, not a square! Also be very careful about the signs of the first two values. If a region occurs under ##x##-axis, then the area is negative; otherwise, it's positive.
     
    Last edited: Jan 9, 2014
  9. Jan 9, 2014 #8
    Ah ok I think I got it:)

    Thanks for your help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Work done by a changing force
Loading...