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Kinetic/Potential/Mechanical Energy Problems

  1. Jan 11, 2012 #1

    k0k

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    A human "cannon ball" in the circus is shot at a speed of 21.0 m/s at an angle of 20 degrees above the horizontal from a platform that is 15.0m above the ground.

    a. If the acrobat has a mass of 56.0 kg, what is his gravitational potential energy relative to the ground when he is at the highest point of his flight? Ignore the effects of air resistance. (Ans: 9.69X10^3 J)

    b. If the net in which he lands is 2.00 m above the ground, how fast is he travelling when he hits it? (Ans: 26.4m/s)

    --
    [Ek=Kinetic Energy, Ep=Potential Energy, Em=Total Energy]
    Ep=mgh
    Ek=1/2mv^2
    W=Fd
    Em=Ep+Ek
    --
    I have found both the Ep and Ek values and added he values up, but the answer isn't anywhere near close. There is something more to it I believe, but I'm not even fully sure how Potiential energy and Kinetic energy are related. I have no idea whether the angle is relevant or not eithier. Explanations would be great help.

    Help would be much appreciated. Thanks~
     
  2. jcsd
  3. Jan 11, 2012 #2
    For a) It's the vertical component of initial velocity that makes sense.
    [tex]0 - v_{0y}^2 = 2( - g)h_0[/tex]
    with h0 is his highest reach relative to the 15-m platform.
    Then the gravitational pe can be deduced easily

    For b) You should also pay attention to the horizontal component [tex]v_{0x}[/tex]. It remains unchange during his flight
    Now you need to find the vertical component [tex]v_y[/tex]
    of the velocity when he hits the net using the law of conservation of energy.
    Finally, add up the two components [tex]v = \sqrt {v_{0x}^2 + v_{_y }^2 }[/tex] and you're done.
    Hope this helps :-)
     
    Last edited: Jan 11, 2012
  4. Jan 11, 2012 #3
    [double post]
     
  5. Jan 12, 2012 #4

    k0k

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    Sorry, but I still don't understand. : /
    How exactly do I start off? What values do I use to calculate mechanical energy then Potential?..
    What about the angle given?..
     
  6. Jan 12, 2012 #5

    NascentOxygen

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    Staff: Mentor

    You start by calculating the vertical & horizontal components of its initial velocity.
     
  7. Jan 12, 2012 #6

    k0k

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    Okay, how does that help me for question a. though?. How do I set up the equation after figuring out the components?.
    (Please bear with me, I'm kinda slow at this.. : / )
     
  8. Jan 12, 2012 #7
    FIY, when he is at his highest point, his velocity is zero, thus his kinetic energy equals zero and his gravitational potential energy is maximum.

    You can either use the kinematic equation: [tex]v_{y}^2 - v_{0y}^2 = 2a(x - x_0 )[\tex]
    where vy=0 and a=-g (vertically, the acrobat undergoes free-falling motion) or use the law of conservation of energy with respect to the platform:[tex]\frac{1}{2}mv_{0y}^2 = mgh_0[\tex]
     
  9. Jan 12, 2012 #8

    k0k

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    Okay, I am using the law of conversation of energy.
    I first look for the total energy from the beginning of the platform using:

    Em= Ep+Ek
    Em=(mgh)+(mv^2 X .5 )
    Em= (56.0kg)(-9.81m/s^2)(15.0m)+ (56.0kg)(21.0m/s^2)(0.5)
    Em= 4107.6J

    Then, I calculate the highest point but since the kinetic energy at the highest point is zero.
    Em=Ep+Ek
    Em-Ek= Ep
    4107.6J - 0= 4107.6J

    I end up getting a potential energy of 4107.6J, not the answer 9.69X10^3 J.
     
    Last edited: Jan 12, 2012
  10. Jan 12, 2012 #9
    Nah, don't get me wrong
    You should always substitute a POSITIVE value for g, say, g=+9.81 m/s^2
     
  11. Jan 12, 2012 #10
    And as we are investigating vertically, v0y= v0*sin(theta)=21*sin20. It should work fine now :-)
     
  12. Jan 12, 2012 #11

    k0k

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    I end up getting 20588.4J instead. : /
    --
    NEVERMIND. I GOT IT. : )
    Thanks
     
    Last edited: Jan 12, 2012
  13. Jan 13, 2012 #12

    NascentOxygen

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    Staff: Mentor

    No, it is not. The body is moving horizontally as well as vertically. Only the vertical component of its velocity is zero at the highest point. So only its vertical K.E. is converted into P.E.

    I think you realized this, but I didn't want to let it slip by without comment.
     
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