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Homework Help: Work-energy theorem problem

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    The masses of the javelin, discus, and shot are 0.8kg, 2kg, 7.2kg, respectively, and record throws in the track events using these objects are about 89 m, 69 m, 21 m. respectively. Neglecting air resistance, (a) calculate the minimum kinetic energies that would produce such throws, and (b) estimate the average force exerted on each object during the throw, assuming that force acts over a distance of 2m.

    2. Relevant equations
    W= [tex]\Delta[/tex]Ek
    W= [tex]\Delta[/tex]Eg
    w = F [tex]\times[/tex] [tex]\Delta[/tex]d
    Eg= mg[tex]\Delta[/tex]h
    Ek = 1/2mv^2

    3. The attempt at a solution
    I attempted to use the law of conservation of mechanical energy, ignoring any effects a non conservative force such as air resistance would have on the system, but am unable to find initial velocity due to the fact that i do not have the height of the projectiles at their maximum point, and i see no way to mathematically remove them from my calculations
  2. jcsd
  3. Mar 11, 2010 #2


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    Write down the equation for the projectile motion. The bodies have horizontal and a vertical velocity components (vx, vy respectively). How these components vary with time? How is related vx a to the length of the shot? How is the time of flight related to the initial vertical velocity?

  4. Mar 11, 2010 #3
    Sorry, but I have never even been though projectile motion, this is a grade 11 physics problem.
  5. Mar 11, 2010 #4
    Projectile Motion is there in Grade 11. You need to check again.
  6. Mar 11, 2010 #5
    Okay, I actually want to learn projectile motion, but in our cricilum in canada we dont do that until grade 12. We did kinematics and dynamics, followed by work, power, and energy
  7. Mar 11, 2010 #6
    This is very easy topic. Do you know about angles and how to resolve them in two perpendicular components?
  8. Mar 11, 2010 #7
    I know about angles, and i did a little bit of reading in to vector components my self, again we do components in grade 12. In this question where not given time, or height, the only possible way I could think of solvign the problem is by persuming the initial velocity, and the angle its being thrown at.
  9. Mar 11, 2010 #8
    I've asked this because I thought telling you something about projectile motion. One think I'm thinking, as kinetic energy and work both are independent of the path taken so there is no need to know projectile motion in this problem. This is only my thinking and I'm not very good in Physics. What do you think about it.
  10. Mar 11, 2010 #9
    But then wouldnt we have to persume the amount of force the athelete applied to the ball.
  11. Mar 12, 2010 #10


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    The motion of these objects (projectiles) can be resolved into a vertical and a horizontal motion, each with an initial velocity. The vertical motion is under the influence of gravity, with acceleration of g. The horizontal velocity [itex]v_x[/itex] is constant. If the object is in air for T time, it travels [itex]D=v_x\cdot T[/itex] horizontal distance. During the same time, it rises to the maximum height and falls back. The rise time is equal to the time of falling down: T/2. If the initial vertical velocity is [itex]v_y[/itex], it will decrease uniformly with rate g and becomes zero at the highest point, in T/2 time:
    [itex]v_y=g \cdot T/2[/itex]

    The initial kinetic energy is

    [tex]KE=0.5 m (v_x^2 +v_y^2 )= 0.5 m (0.25 g^2\cdot T^2+D^2/T^2) [/tex]

    and you want it to be minimum.

    Local monimum can exist if the derivative with respect to T is zero, that means

    [tex]0.5 \cdot g^2\cdot T-2\cdot D^2/T^3 = 0 \rightarrow T^4=4D^2/g^2 \rightarrow T=\sqrt{2D/g}[/tex].

    This means that initially vx should be equal to vy, to cover a distance with the lowest energy.

    The kinetic energy of the object becomes

    [tex]KE= 0.5 m \cdot D\cdot g [/tex].

    According to the work-energy theorem, so much work had to be done on it. Work is force times distance along it acts, 2 m in our case, so

    [tex]F = 0.25*m*D*g [/tex] .

    Last edited: Mar 12, 2010
  12. Mar 12, 2010 #11
    ehild. John is grade 11. calculus is not an option in canada 'til grade 12 (unless you are part of the AP/IB curriculum, in which you do vectors in grade 11 as well.) However, you do prove a good point that can be logically reasoned out without calculus.

    "This means that initially vx should be equal to vy, to cover a distance with the lowest energy."

    Meaning the throwing angle must be 45 deg. With this angle one can resolve the question quite neatly as you follow in your analysis.
  13. Mar 12, 2010 #12


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    Well, the 45° angle as most effective to throw something at farthest distance might be mentioned in John class. I derived it for the sake of the other possible readers.

  14. Mar 12, 2010 #13
    ehild, I think you must use latex code because it is very hard to read and understand.
  15. Mar 12, 2010 #14

    As you can see in the image above that [tex]v_0[/tex] is resolved into two rectangular component [tex]v_x[/tex] and [tex]v_{0y}[/tex] and the trajectory of any body in this case is parabola.

    Now you can see that the acceleration along x axis is zero and on y axis the acceleration due to gravity is acting downward.

    Now, as you are new to this topic I must tell you what happens with the objects' velocity when you throw it up in the air. The vertical velocity continuously decrease with the rise in height due to acceleration due to gravity and then after some time the velocity become zero and then particle start falling downward.

    So when object is at his maximum height the object's vertical velocity is zero.

    use equation of motion to find maximum height.

    v = 0 [tex] u = v_{0}sin\theta[/tex] and a = -g

    [tex] v^2 = u^2 + 2aH[/tex]
    [tex] 0 = u^{2}sin^{2}\theta - 2gH[/tex]
    [tex] H = \frac{u^{2}sin^{2}\theta}{2g}[/tex]

    time taken to reach that maximum height can also be calculated

    [tex] v = u + at[/tex]
    [tex] 0 = u sin\theta - gt[/tex]
    [tex] t = \frac{u sin\theta}{g}[/tex]

    as time of ascent is equal to time of descent

    Time of flight = Time of ascent + Time of descent

    [tex]T = \frac{2u sin\theta}{g}[/tex]

    and Range = horizontal velocity multiplied of Time of flight

    [tex] R = u cos\theta \frac{2u sin\theta}{g}[/tex]
    [tex]R = \frac{u^2 Sin 2\theta}{g}[/tex]
    Last edited by a moderator: Apr 24, 2017
  16. Dec 10, 2010 #15
    to snshusat161, what does the u mean in your equations?
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