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What is the work done in this case?

  1. Apr 20, 2010 #1
    What is the work done in this case??

    1. The problem statement, all variables and given/known data
    A car of mass 1000 Kg moves with velocity 2 m/s , the brakes are applied to stop the car after 2 sec . Calculate the work done by the brakes.



    2. Relevant equations

    F=ma

    Vt-V0=at
    W = F*d*cos theta
    d = V0t+1/2 a t^2
    3. The attempt at a solution


    F=ma

    Vt-V0=at
    0-2=2a
    Then a = -1 m/s^2
    k..
    F = -1000 N

    d = V0t+1/2 a t^2
    d = 4 + (0.5*-1*4)=2m

    since the force applied by the brakes is opposite to the displacement
    then theta = 180 degrees

    W = F*d*cos theta
    W= -1000*2*-1 = 2000 joule

    I'm sure that there's something wrong with my answer (to be more exact I think I've somethi wrong with the signs)

    So could you help?
    Thanks
     
  2. jcsd
  3. Apr 20, 2010 #2
    Re: What is the work done in this case??

    You were very close.

    [tex]W=\vec F\cdot \vec d[/tex]
    [tex]W=|F| |d| \cos{\theta}[/tex]

    You said that the magnitude of the force is negative, but that is untrue! The information that the force is opposite the displacement is already in the cosine term, putting the minus in twice is a mistake. :)

    Either remember the vector formula, or the one with the absolute values to help avoid the confusion!

    Another way of approaching the problem would be to use the work energy theorem which states that the work performed on an object is equal to the change in its kinetic energy. In this case, the final kinetic energy is 0, and the initial kinetic energy is 2000 joules. The work equals the final kinetic energy minus the initial kinetic energy, which is -2000 joules.

    In its general form, when no other forces are present but the one doing work, the work energy theorem states:
    [tex]W=\Delta K = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2[/tex]
     
  4. Apr 20, 2010 #3
    Re: What is the work done in this case??

    Great!
    How bout the negative acceleration ?How can i consider it in the problem?
    Do u mean that we just use the magnitude of force (and although it is negative we just use its magnitude)?
     
  5. Apr 20, 2010 #4
    Re: What is the work done in this case??

    Exactly. :) Whether the force is negative or positive (Along or opposite the displacement is what matters, not your coordinate system!), is expressed in the formula in the cosine of the angle.
    The magnitude of a vector can only ever be a positive quantity, remember that, and you will avoid a lot of confusion.
     
  6. Apr 20, 2010 #5

    ideasrule

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    Re: What is the work done in this case??

    Why are we doing all of this? The work done by the brakes is just the kinetic energy of the car.
     
  7. Apr 21, 2010 #6
    Re: What is the work done in this case??

    I got it .
    Thanks for all of you
     
  8. Apr 21, 2010 #7
    Re: What is the work done in this case??

    Sometimes, when people haven't learned about kinetic energy, and are still working on forces, teachers don't usually assign work and expect students to do the problem with the energy.

    Still, you're right, it's abour 70 times easier considering only energy.
     
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