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, Energy and Momentum

  1. Oct 10, 2005 #1


    User Avatar

    I need help with this problem...please help me. I'm sorry the axis of the graph look a little weird...my scanner is not working so I have to make the graph myself T___T


    A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object's displacement as shown in the graph above. The object starts from rest at displacement x=0 and t=0 and is displaced a distance of 20m. Determine each of the following.

    a. The acceleration of the particle when it's displacement x is 6m

    b. The time taken for the object to be displaced the first 12

    c. The amount of work done by the net force in displacing the object the first 12m

    For this one, I know W=Fx so I just take W= 4(12)= 48 J

    d. The speed of the object at the displacement x=12

    e.The final speed of the object at displacemnt x=20 m

    f. The change in momentum of the object as it is displaced from x=12m to x=20m

    I think I can work this out by using p= mv*final* - mv *initial*

    For the whole problem..well, I know the area of the graph will be work...and W=Fx but I still don't know how it's related to acceleration or speed though? I'm just confuse with how to find the acceleration and speed. Once I have figure it out I think I'll know how to do the rest of the problem. So please help me just a little bit ^_^
    Last edited: Oct 10, 2005
  2. jcsd
  3. Oct 10, 2005 #2
    Hi M&M

    The acceleration is given by the second law [tex]a=\frac{F}{m}[/tex]. You can divide the F(x) graph by m and you obtain a(x).

    As for velocity you can apply [tex]v=\sqrt{2ax}[/tex] (Galilei) until x=12 m. Beyond this limit the acceleration is decreasing linearly so you can use the same Galilei's law with an average acceleration:
    Then the velocity in any point at [tex]x>12m[/tex] is given by
    [tex]v^2(x)=v^2(12m)+2a_m (x-12m)[/tex]
    Last edited: Oct 10, 2005
  4. Oct 11, 2005 #3


    User Avatar

    Thanks a lot for answering my question..I figure out all of them except for

    e. which is:

    The final speed of the object at displacemnt x=20 m

    So if I want to find the acceleration when x=20, I take:

    a= F/m and since F=0 then a= 0 too

    then I plug it into the last equation and solve for v but since a = 0 then v= 0 too right since the object start from rest?!?! I hope you understand what I mean (_ _ ")

    I always have a feeling that it's wrong when my answer is 0 ^_^
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Work, Energy and Momentum
  1. Work Energy Momentum (Replies: 5)