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!

Problem: Work done by a Variable Force

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    An object is acted on by the force shown below. What is the final position of the object if
    its initial position is x = 0.40m and the work done on it is equal to
    (a) 0.21 J
    (b) (b) -0.19 J.

    [​IMG]

    2. Relevant equations
    W = F1x1 + F2(x2 - x1)


    3. The attempt at a solution

    a) W= F(x2-x1) = 0.8N* (x2- 0.40m)

    but i DON'T get the answer correctly.
     
    Last edited: Oct 23, 2008
  2. jcsd
  3. Oct 23, 2008 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    How do you determine the work done from the graph you have given? Does that correspond to the formula you gave?
     
  4. Oct 23, 2008 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi lucky_star! Welcome to PF! :smile:

    Why have you used only one value of F?

    F has different values as x changes.
     
  5. Oct 23, 2008 #4
    Oh, I made a mistake there. if initial position is 0.40m, the force is 0.8N. I am given W= 0.21J, So I will have:

    0.21J= 0.8N*0.40m + 0.8N*(x2 - 0.40m)
    0.21J= 0.8x2 - 0.32
    x2= 0.66m

    That's all I have. However, the correct answer that I have been provided is 0.90m. I really don't know how to solve this problem. Please give a hint!
     
  6. Oct 23, 2008 #5
    Re: Welcome to PF!

    I know the initial position is .40m, but I don't know which are the corresponding forces :(
     
  7. Oct 23, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Noooo … :cry:

    Look at the graph!

    The force is 0.8N only for x ≤ 0.50m.

    Try again. :smile:
     
  8. Oct 23, 2008 #7
    AHHH! I Think I understand the problem.

    W= 0.8N *(0.5m-0.4m) + 0.4N *(0.75m-0.5m) + 0.2 *(x2-0.75m)
    0.21J = 0.080+ 0.1 + 0.2x2
    0.18 = 0.2x2
    x2 = 0.90m
     
  9. Oct 23, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin: Woohoo! :biggrin:

    Have a free quark! o:)

    (and now try b … :wink:)
     
  10. Oct 23, 2008 #9
    do I just simply plug in the value of w= -0.19J and then solve it, or else?
     
  11. Oct 23, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    You tell me! :wink:
     
  12. Oct 23, 2008 #11
    When I substitute -0.19J in the equation I did not get the right answer. thus, I think it's not a substitution, because work have a negative sign right now. So, i am not sure what to do :(


    p/S: I have class in about 20 more minutes, and this is my last question.
     
  13. Oct 23, 2008 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hint: work = force x distance,

    so if work is negative, then … ? :smile:
     
  14. Oct 23, 2008 #13
    Ok. I got to run to class now, I'll try in class. Thank you so much for helping me. Have a nice day.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Problem: Work done by a Variable Force
Loading...