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Need Help with Graph Question

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    This Question has a graph with it. (See Attachment)

    An object of mass (m) moves in a straight line. The net force varies with the displacement. The mass starts at 0 meters and at 0 seconds.The mass goes a total distance of 20 meters. X1 is equal to 11 meters.

    Mass --- .87kg
    Net Force --- 4 Newtons
    X1 --- 11 meters

    Find the:
    Acceleration of the object when its displacement is 11/3 meters?
    The time taken for the object to be displaced from 0 to X1 (11 meters)?
    Amount of work done by the net force in displacing the object from 0 to X1 (11 meters)?
    The speed of the object at displacement X1 (11 meters)?
    The final speed of the object at 20 meters?
    The impulse experienced by the object as it is displaced from X1 to 20 meters (9 meters)?

    2. Relevant equations

    F=ma
    Vf^2=Vo^2+2as
    W=Fd

    3. The attempt at a solution

    I am not sure if I am doing this right, can someone make sure?

    For the acceleration would it be ---- F=ma so --- 4N=.87kg(a) ---- a= 4.597m/sec

    For the speed of the object at displacement X1 (11 meters) I used
    Vf^2=Vo^2+2as ----- Vf^2=(0)+2(4.597m/s)(11meters) ---- Vf= 10.1 m/s

    For the time I used --- t=d/(average velocity) ---- Average velocity= (initial+final/2) =
    (0+10.1/2) = 5.05 m/s ? (I am not sure about this).... Then I put it into t=d/(average velocity) ----- t=(11 meters)/(5.05m/s) = 2.19 seconds? (check please)

    For the work W=Fd ---- W=(4)(11) ---- 44J

    I am not sure how to get the overall final speed of the mass because the force is changing at the end.

    I have no idea how to do the impulse experienced by the object from X1 (11 meters) to 20 meters, I am not sure what the equation is.

    Can you tell me if I am doing this correctly and how I should try to get the last two parts.

    [​IMG]
     

    Attached Files:

    Last edited: Nov 23, 2007
  2. jcsd
  3. Nov 23, 2007 #2
    Cannot see the picture.It says "Atachment pending approval".
     
  4. Nov 23, 2007 #3
    I have the graph up now.
     
  5. Nov 23, 2007 #4
    For time
    [tex]x=v_0t+\frac{1}{2}at^2[/tex]

    [tex]t=\sqrt{\frac{2x}{a}[/tex]

    To find overall velocity:
    Calculate the area under the graph.[tex]Area=\frac{1}{2}mv_f^2[/tex]

    Impulse:

    [tex]I=m(v_f-v_{x_1})[/tex]

    [tex]v_f[/tex] is the velocity at x=20 m

    [tex]v_{x_1}[/tex] is the velocity at x=11 m
     
  6. Nov 23, 2007 #5
    Other solutions seems to be right.
     
  7. Nov 24, 2007 #6
    I do not understand how the area under the graph could get me the final velocity, can anyone explain?
     
  8. Nov 24, 2007 #7
    You need to find to find this area in yellow.

    [​IMG]
     
  9. Nov 24, 2007 #8
    Because area under the F-x graph equals to work.Here work goes only for kinetic enrgy.
     
  10. Nov 25, 2007 #9
    So is the final velocity 11.9 m/s
     
  11. Nov 25, 2007 #10
    Yes,you're right!!
     
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