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Math Help - Baseball - kinetic energy

  1. #1
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    Baseball - kinetic energy

    I can't even figure out where to begin. Here's the way the problem reads:


    In this problem we calculate the work required for a pitcher to throw a 90-mi/h fastball by first considering kinetic energy.

    The kinetic energy of an object of mass and velocity is given by .

    Suppose an object of mass , moving in a straight line, is acted on by a force that depends on its position . According to Newton’s Second Law



    where and denote the acceleration and velocity of the object.



    (a) Show that the work done in moving the object from a position to a position is equal to the change in the object’s kinetic energy; that is show that



    Where and are the velocities of the object at the positions and . Hint: By the Chain Rule,





    (b) How many foot-pounds of work does it take to throw a baseball at a speed of 90 mi/h?
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  2. #2
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    Quote Originally Posted by ReneePatt View Post
    I can't even figure out where to begin. Here's the way the problem reads:


    In this problem we calculate the work required for a pitcher to throw a 90-mi/h fastball by first considering kinetic energy.

    The kinetic energy of an object of mass and velocity is given by .

    Suppose an object of mass , moving in a straight line, is acted on by a force that depends on its position . According to Newton’s Second Law



    where and denote the acceleration and velocity of the object.



    (a) Show that the work done in moving the object from a position to a position is equal to the change in the object’s kinetic energy; that is show that



    Where and are the velocities of the object at the positions and . Hint: By the Chain Rule,





    (b) How many foot-pounds of work does it take to throw a baseball at a speed of 90 mi/h?
    For part a) start with Newton's law:

     F(s) = m \frac{dv}{dt}

    Now multiply both sides by  ds .

     F(s) \, ds = m \frac{dv}{dt} ds

    Now, note that  \frac{dv}{dt}ds = \frac{ds}{dt} dv = v dv

    Hence

     F(s) \, ds = m v dv

    Not integrate both sides:

     \int_{s_1}^{s_2} F(s) ds = \int_{v_1}^{v_2} mvdv

    For part b), you are simply subbing numbers into the equation that you derive in part a).
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  3. #3
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    So for part b - where do I sub in the number?

    Quote Originally Posted by Mush View Post
    For part b), you are simply subbing numbers into the equation that you derive in part a).
    Where do I sub in the numbers for part b?
    I don't understand. How do you find the foot-pounds of work to throw the ball at a speed of 90 mi/h?
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