1. Limited time only! Sign up for a free 30min personal 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!

How much work is done?

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Question
    A 7.6 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F with arrow acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in the figure. The force F with arrow is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F with arrow between t = 0 and t = 0.5 s?

    7-p-007.gif

    2. Relevant equations
    x=vot +1/2at^2
    v=at
    m/2(Vf^2-Vi^2)
    3. The attempt at a solution
    When I work the problem I get 0.152 Joules.
    .8 = 1/2 a(2.0)^2
    a=0.4
    v=0.4(0.5)
    v=0.2
    1/2(7.6kg)(.2^2)= 0.152J.

    I work it this way and I have no idea why it is wrong. The answer is supposed to be 0.0973.
     
    Last edited: Dec 7, 2015
  2. jcsd
  3. Dec 7, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Where does .8 come from?
    The problem statement is inconsistent - a constant force cannot produce this pattern. There are multiple ways to approach this problem and they will lead to different answers.
     
  4. Dec 7, 2015 #3
    plug it into 1/2 at^2 = x along with t = 2.0 to get the acceleration. Then I use that to find the velocity at .5 seconds.
     
  5. Dec 7, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I know, but where does the number 0.8 come from? It is not the position of the object after 0.5 seconds.
    Edit: wait, you used the 2 seconds. I guess that's the origin of the discrepancy.

    Edit2: Do the same calculation with 0.5 seconds and you get half the other value. Weird.
     
  6. Dec 7, 2015 #5
    Ah... yea thanks... though something is off with that problem that doesn't make sense... got it to work now though.
     
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: How much work is done?
  1. How much work is done (Replies: 8)

Loading...