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Homework problem- Work and energy problem

  1. Jun 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A 68 kg in-line skater starts from rest and accelerates at 0.21 m/s square for 15s.

    a) Find her final velocty and total kinetic energy after the 15s of travel.

    b) If she exerts a breaking frictional force of 280N, find her stopping distance.

    2. Relevant equations

    Ek=1/2mv square
    Eg=mg triangle (height)- triangle=multiply
    Wnc=Efinal-Einital

    3. The attempt at a solution

    I got most of the equation i knw. I need help solving it with a step by step solution.
     
    Last edited: Jun 1, 2007
  2. jcsd
  3. Jun 1, 2007 #2

    danago

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    Think about what acceleration means. 0.21m/s2 means that for every second she travels, her speed will increase by 0.21 m/s. So what final velocity results from travelling for 15 seconds?
     
  4. Jun 1, 2007 #3
    so 3.15 rounded to 3.20 sound right. I also need to knw the step by step solution for total kinetic energy after 15 secs. and If she exerts a breaking frictional force of 280 N, find her stopping distance.
     
  5. Jun 1, 2007 #4

    danago

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    Well consider the formula for kinetic energy:

    [tex]
    E_k = \frac{{mv^2 }}{2}
    [/tex]

    You are given the mass of the skater, you have calculated the velocity after 15 seconds, so you can easily calculate the kinetic energy by using the formula.
     
  6. Jun 1, 2007 #5
    oh i got the kinetic energy of 3.4 X 10 to the power of 2 J. Thanks. For question b.) if she has a frictional force of 280N, how would i solve the stopping distance?
     
  7. Jun 1, 2007 #6

    danago

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    If she is slowing down, her velocity is changing, thus, she is accelerating (or decelerating in this case). Have a look at newtons laws, and see if you can figure out how to find what deceleration the force causes.
     
  8. Jun 1, 2007 #7
    I found F=deltra (mv)/time but i dun tink i got the right answer.
     
  9. Jun 1, 2007 #8

    danago

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    For an acceleration to occur, a net force must be present. This is represented by the formula:

    F=ma

    This pretty much says that if a net force (F) acts on a body, it will accelerate. The actual amount of acceleration depends on the mass of the body.

    In this case, you are told the 280N is a frictional force, therfore, it is acting against the skaters velocity, and slowing her down. Using the given force and mass, you can calculate the deceleration the force causes. With that, have a look at your equations of motion and see if you can figure the rest out.
     
  10. Jun 1, 2007 #9
    I got an answer of 4.1 for acceleration but the question states. If she exerts a breaking frictional force of 280N, find her stopping distance. Is 4.1m her stopping distance?
     
  11. Jun 1, 2007 #10

    danago

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    No. 4.1 is the deceleration. It means that for every second that passes, her velocity decreases by 4.1 m/s.

    The next step is to calculate what distance she travels while her velocity is going from 3.15 to 0, with a deceleration of 4.1. Have you been given a set of formulas that you can solve this with?
     
  12. Jun 1, 2007 #11

    danago

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    Have you been given these forumulas? :

    [tex]
    \begin{array}{l}
    s = ut + \frac{{at^2 }}{2} \\
    v^2 = u^2 + 2as \\
    \end{array}
    [/tex]
     
  13. Jun 1, 2007 #12
    Nope i didnt receive or use those formulas before. So how would i find the distance?
     
  14. Jun 1, 2007 #13

    danago

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    Well consider the equation W=Fs.

    The work (W, or change in energy) is equal to the force multiplied by the distance over which the force acts.

    The change in kinetic energy will be the initial kinetic energy (which you worked out in part (a)), take the final kinetic energy, which is when v=0. Use this, aslong with F=280, to find the distance over which the forcer acts, which is the stopping distance.
     
  15. Jun 2, 2007 #14
    oh ic. Thanks alot. I figured out the answer. W=Fs
    W=3.4X10 to the power of 2
    F=280
    s=?

    so W=Fs
    340=280s
    s=1.2m

    Therefore the stopping distance is 1.2m.
     
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