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Calculate kinetic energy transferred

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/hiNHMZZ.jpg - I am struggling with the right-hand page, questions b i, b ii and b iii. I am not sure how to calculate the total amount of kinetic energy transferred to the fluid when given the mass before and after the experiment?

    2. Relevant equations
    EK = 1/2 x m x v^2

    3. The attempt at a solution
    I was thinking maybe you use 7.5 ms^1 and work out the amount of KE before and after the experiment, then find the difference. I am doubting this because the fluid left the bottle at 7.5 ms^1 and I'm not sure if this is relevant to the final velocity of the fluid.
     
  2. jcsd
  3. Nov 26, 2014 #2

    BvU

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    Well, first you want to check the lower left: is that really the horizontal distance travelled ?

    Then, for the right-hand page: before the experiment, KE is zero. Afterwards everything is wet but the liquid doesn't move any more, zo zero again. That's not what they mean: before and after are provided to help you calculate how much mass has left the bottle. And it 'all' left the bottle at this 7.5 m/s if you want to believe the story. But then, ii) and iii) give you the opportunity to comment on that.
     
  4. Nov 26, 2014 #3

    gneill

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    Hi Nicaragua.

    First of all, you should not use rounded values of intermediate values. That is, if you calculate some value in one step and then need that value in a subsequent step, don't use its rounded value. Instead, keep several extra decimal places as "guard digits" to prevent rounding error from creeping into your significant figures. Round values only for presentation of results.

    Next, I notice that you have not calculated the maximum horizontal distance correctly. Since the fluid comes out of the bottle at a given speed there is no acceleration taking place from 0 m/s up to that speed over the trajectory. It's projectile motion with an initial velocity. So, choose another formula for determining the time of the trajectory, or look up and use the "range equation". Also note that the horizontal speed is not 7.5 m/s. You calculated the horizontal component of the velocity in a previous step.

    For the kinetic energy calculation you make an assumption that the total mass of the ejected fluid was given the same speed, namely 7.5 m/s. Apply the KE formula with that mass and velocity.
     
  5. Nov 26, 2014 #4
    Thanks very much for the advice. So for calculating the range, which equation of motion would I use? If I only have the values u, a and v of the vertical component, then I can only think to use v = u + at?

    So for question b, I have used each different mass in the EK equation and found that before the experiment it had 0 kinetic energy (because velocity is zero) and afterwards it had 22.219 Joules.
     
  6. Nov 26, 2014 #5
    It says the fluid left the bottle at a speed of 7.5 ms^-1 ? What equation should I have used? Thanks
     
  7. Nov 26, 2014 #6

    gneill

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    What's the equation of motion for the position of an object that has an initial velocity and undergoes constant acceleration? You want to deal with the vertical component of the motion alone at first.

    And as I mentioned before, as an alternate approach you could look up the "Range Equation" .

    Show your work.
     
  8. Nov 26, 2014 #7
    I thought I was dealing with the vertical component alone first? In terms of the vertical component I have initial velocity, acceleration and want to know time, so used the equation v = u +at? Also I can't use the range equation as my tutor wants this to be calculated using EOM.

    And to work out the EK I used the formula EK = 1/2 m v^2 , plugged in the values to get KE = 1/2 x 0.79 x 7.5^2 = 22.219 Joules.
     
  9. Nov 26, 2014 #8

    gneill

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    Okay. What values will you put in for v and u? Remember that velocity is a vector and has both magnitude and direction. Show your calculation.
    0.79 kg is the final mass of the bottle+remaining contents. That's not the mass that was ejected...
     
  10. Nov 26, 2014 #9
    For v I will put the value as 0 m/s because thats when the liquid will hit the ground/ be at its peak. For u I will put 6 m/s because this is what I found out previously.

    And for the kinetic energy, can you give me some more help because I am really struggling for ideas.
     
  11. Nov 26, 2014 #10

    gneill

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    0 m/s is the speed of the water after it hits the ground and has lost all its kinetic energy. So you don't want to use that for v. For u, don't use rounded intermediate values! You'll lose accuracy in your calculations, eventually getting significantly wrong answers. You found 5.7 m/s for the initial y-speed, then rounded that up to 6 m/s. Use the un-rounded value. In fact, you should use a few more digits: 7.5 cos(50°) = 5.745 would be a better value to use to avoid rounding error getting into your significant figures. Youcan round results when you're finished calculating.

    Now, what is the speed of the water in the instant just prior to it landing? What sign will you give that speed to indicate its direction (velocity is a vector quantity).
    How much water left the bottle?
     
  12. Nov 26, 2014 #11
    So what quantities of the vertical component do I have? Acceleration, initial velocity and that's it?

    As for question b, it doesn't say how much water is left, only the total mass before and after. Can you please tell me how I go about this? I appreciate your help in trying to make me figure it out but I still only know as much as when I started. Thanks
     
  13. Nov 26, 2014 #12

    gneill

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    By conservation of energy (kinetic energy versus gravitational potential energy), the speed at landing must equal the speed at launch. Only the direction differs.
    Well, if you know the mass of the container before, and you know the mass of the container after, isn't the difference of those two equal to what was removed from the container?
     
  14. Nov 26, 2014 #13
    So in terms of vertical components, does u=v? I might have been mixing it up with the horizontal thinking v=u. Anyway, what equation would even solve this?

    Ahh I see. So the liquid has a mass of 1.45kg. 1/2 x 1.45 x 7.5^2 = 40.781 Joules
     
  15. Nov 26, 2014 #14

    gneill

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    The equation you've chosen is fine for finding the time so long as the correct values of u and v are used. As stated previously, the magnitudes of u and v are the same thanks to conservation of energy, but the directions are different. At launch the vertical direction of motion is upwards. At landing the vertical direction of motion is downwards. So sort out the signs of u and v according to your choice of coordinate system.
    Yes.
     
  16. Nov 26, 2014 #15
    Oh ok so I would still use v = u + at? Just v = -5.7 m/s and u = 5.7 m/s ?

    Thanks for question b, I totally get it now.
     
  17. Nov 26, 2014 #16

    gneill

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    Why don't you make a sketch of a projectile following the same path and draw in the velocities and their components at the launch and landing, also include the direction of the acceleration vector. Then convince yourself that the above values make sense.

    Once you've done that, do the calculation using the values you've settled on and present your value of t.

    Again, I suggest that you use more guard digits in your intermediate values.
     
  18. Nov 26, 2014 #17
    Please can you take me through how you would solve the range question and I can ask any questions afterwards? I am just so confused now...with every reply I feel more and more lost. It would really help me see if you take me through the question. Thanks man
     
  19. Nov 26, 2014 #18

    gneill

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    Sorry, that's against the rules. We can't do your work for you. We can only offer suggestions, hints, corrections, and so on.

    I can tell you that once you have the "air time" of the projectile (the time from launch until it hits the ground then that tells you the horizontal traveling time, too. You've got a formula that will work for you to find the time. All you need to do is sort out the values (and their signs) to plug in for u, v, and a. You're almost there. It's important that you make the sketch and really understand where the values come from and why they have the signs they do. Once you do that you'll always be able to sort these things out via a quick sketch.
     
  20. Nov 26, 2014 #19
    I don't understand what you mean. I just drew a sketch and know the initial horizontal and vertical velocities, and i pres-sume the final vertical velocity is -5.7 m/s. No wait, it should be 0 m/s? Ahhh this is so frustrating that you're not allowed to be clear with me.
     
  21. Nov 26, 2014 #20

    gneill

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    If upwards is the positive direction of your vertical axis then the initial velocity is +5.745 m/s. The final velocity is in the opposite direction, so it's -5.745 m/s. You should see that on your sketch. Now, what's the value of the acceleration a? What direction does it point on your axes?
     
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