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Physics help on force and speed

  1. Jul 31, 2013 #1
    1. The problem statement, all variables and given/known data

    FIGURE 7-6




    The force on a 3.00-kg object as a function of position is shown in Fig. 7-6. If an object is moving at 2.50 m/s when it is located at x = 2.00 m, what will its speed be when it reaches x = 8.00 m?

    phpkV1GdB.png
    ________
    A)

    2.90 m/s

    B)

    3.30 m/s

    C)

    3.70 m/s

    D)

    4.10 m/s

    E)

    4.50 m/s


    2. Relevant equations

    Have no idea

    3. The attempt at a solution
    the answer is 3.3 m/s but I don't know how they got that answer
     
    Last edited by a moderator: May 9, 2015
  2. jcsd
  3. Jul 31, 2013 #2

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    You might want to use work energy methods and show an attempt, please. Thank you.
     
  4. Jul 31, 2013 #3
    I tried but my answer came out to be something that's not even on the answer choice lol
    Here is what I did

    Δk = w
    1/2 mv^2 = Fd
    (1/2)(3)v^2 = 1*8
    V= 2.3 m/s
     
  5. Jul 31, 2013 #4
    I got your answer as follows.
    First find an expression for the acceleration vs distance. Then integrate the acceleration to get the velocity vs distance. Use the fact that at x = 2 the velocity = 2.5 to find the constant of integration. Now you can find the velocity at x = 3. Now, knowing the velocity, mass and force at x = 3, you can find the velocity at x = 8.
     
  6. Jul 31, 2013 #5
    You had the right idea, but you made a couple of mistakes. When you wrote the change in kinetic energy on the left hand side of the equation, you left out the term for the initial kinetic energy at x = 2. On the right hand side of the equation, you obtained the wrong value for the integral of the force with respect to x, from x = 2 to x = 8. The correct value should be 6.5.

    Chet
     
  7. Jul 31, 2013 #6
    Got it!! Thank y'all :)
     
  8. Oct 17, 2016 #7
    • Member advised not to provide complete solutions to homework questions.
    Change in K = W
    1/2*m*(vf^2-v0^2) = F*d
    .5*3.00kg*(vf^2-(2.5m/s)^2) = 1N * (8-2m)
    1.5 (vf^2-(2.5m/s)^2) = 6N*m
    vf^2 = 6/1.5 + (2.5 m/s)^2
    vf^2 = 10.25
    vf = 3.20 m/s
     
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