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Homework Help: Work-kinetic energy theorem

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi all, I am currently working on a problem that i believe involves the work-kinetic energy theorem.
    See the attachment and information below.

    As shown in the figure, the geometry of the water slide used is 1m:
    Slide length = 61,; Slope angle = 24; Ramp height = 3.66 m

    Lubricant is used thus friction can be ignored. A man of 70kg slides from the start.
    if the ramp angle is 30, find the magnitude and direction of the velocity of this man when he just flies away from the ram and the Horizontal distance that the man flies.

    2. Relevant equations

    W=(1/2)mvf^(2) - (1/2)mvi^(2)

    3. The attempt at a solution

    ok so first I find the force that is acting on the man as he starts sliding down the ramp. which comes out to be mgcos(24) = F

    and i know the distance of the slide which is 200ft. And further we can say W=FD
    so subbing values in and solving for vf I get 108.43 ft/s - this is the velocity when the man is at the the lowest point on the slide.

    So then i find i repeat the process and find vf at the top of the ramp with an angle of 30. which gives me an answer of 102.08 ft/s.

    There is no answer to this question so i have no idea if i'm right or not.... any help would be greatly appreciated.

    also to find the direction of this velocity can i assume at the instant he comes off the slide then the angle the slide makes to the horizontal is also the direction of his velocity?

    Thank you again!

    Attached Files:

  2. jcsd
  3. Apr 12, 2013 #2
    disregard my initial workings, i used cosine when i should have used sine... still not sure if my answers are correct however.

    My new answer is 66.8 ft/s when the man leaves the ramp.
  4. Apr 12, 2013 #3

    Simon Bridge

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    I'm afraid that docx file won't come out on my computer.
    Please do not provide important information in files that contain proprietary formatting and/or extensions - not everyone can open them.

    From your description, this is a conservation of energy problem but you don't need the work-energy theorem.
    Start by describing the motion of the man in terms of the energy changes that occur in each stage.
  5. Apr 12, 2013 #4
  6. Apr 12, 2013 #5

    Simon Bridge

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    OK - now I get it.

    According to that diagram, the distance from the top of the slide to the point on the slope directly below the top of the ramp is 200 feet. The man does not go down a 200 foot slide and then up a ramp ... the ramp is part of the 200 feet and the overall length of the (colored brown) surface is going to be greater than 200 feet overall.

    So the work-energy relation is not going to help you here - the force you worked out does not act for the entire 200 feet. *

    My suggestion at post #3 stands: describe the motion in terms of energy transformations and use conservation of energy.

    What sort of energy is lost going from the top of the slide to the top of the ramp?
    What does it turn into?


    [*] actually the arrow indicating the 200 foot length stops a bit short - yet goes past the curve to the bottom of the ramp ... so it is a little ambiguous. You could check what is intended here - but, if you use conservation of energy as suggested, it won't matter ;)
    Last edited: Apr 12, 2013
  7. Apr 13, 2013 #6
    Ok so my new attempt is as follows:

    mgh = (1/2)*mv^(2)

    h= 69.35 ft - subbing this into the equation and solving for v i get 66.8 ft/s.
    Not terribly confident however that this is correct because its the same result i got using an incorrect method.
  8. Apr 13, 2013 #7

    Simon Bridge

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    You left out a lot of steps there so I cannot see how you got those numbers.

    You did: h=200(ft)sin(24°)-12(feet)=69.347 ft
    ... with g=33ft/s/s, v=√(2x33x69.347) = 67.653ft/s

    Perhaps several errors cancelled out?
    Perhaps I misread what you did?
    Whatever - can you see how this way was more direct?
  9. Apr 13, 2013 #8
    yes, however there is a second part to the question which is also troubling me. the question also requires me to find the direction of this velocity when he leaves that ramp. can i assume it will be the same angle that the ramp makes with the horizontal i.e 30?

    I will need this value to find out the horizontal distance that the man flies when he leaves the ramp.
    Last edited: Apr 13, 2013
  10. Apr 13, 2013 #9

    Simon Bridge

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    That's correct.
    After that - finding the horizontal distance is normal ballistics.
  11. Apr 13, 2013 #10
    Hi Simon - sorry to pester you again but can i get you to verify my answer as there is no solution i have no idea if my working is correct :(.

    Ok so my solution to find the horizontal distance is as follows:

    finding the final y component of velocity
    Vfy^(2) =Viy^(2) +2a*Δy
    I have the variables to solve for Vf, since Vi^(2) = 66.8sin(30), g = -32.174 and Δy = -12
    so Vf = 43.44 ft/s

    Now i use this to find time.

    Vfy = Viy +at

    So Vfy = 43.44 Viy = 66.8sin(30) and a = -32.174

    so t = 2.38s

    and thus Δx can be found.

    Δx = Vix * T = 66.8cos(30)*2.38 = 137.683 ft

    Thanks again!
  12. Apr 13, 2013 #11
    Ok so the final part of the question is as follows:

    Use theoretical methods to determine the magnitude of the ramp angle that will result in the max flying distance.

    Honestly I don't know where to start with this question. If i try and relate the potential energy to the kinetic energy the terms in the equation don't involve the ramp angle and thus is unhelpful. any tips would be greatly appreciated.

  13. Apr 13, 2013 #12

    Simon Bridge

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    In #10 you did it in several steps using several equations - use algebra to combine the equations until you have only the initial velocity, gravity, and an angle.

    I think your teacher is trying to get you out of the habit of thinking in terms of equations you have memorized and start you thinking in terms of constructing the equations you need.
  14. Apr 13, 2013 #13
    Ok so,

    Δx = Vocos(∅)t

    Δy = Vosin(∅)t -(1/2)gt^(2)

    since Δy =0, we can solve for t and sub it into the first equation and get an equation in terms of gravity, intial velocity and the angle.

    doing this yields (2Vo^(2)sin(2∅)) / g

    and this is the expression i need to maximize?
  15. Apr 13, 2013 #14
    actually disregard that i made a mistake
  16. Apr 14, 2013 #15

    Simon Bridge

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    Yep - i.e. ##\Delta y \neq 0##
    But I suspect you already know the angle that maximizes the flying distance (with no air resistance) - you just have to prove it.

    It helps to simplify the notation ... let the initial speed be v so that the horizontal speed is ##v_x=v\cos\theta##
    Let the horizontal distance traveled by d so that ##d=v_xt## where t is the time traveled. The vertical distance travelled is the height of the ramp - h.

    The initial vertical speed is ##v_y=v\sin\theta##

    This lets you avoid all those subscripts and deltas that can mess with your algebra.
    (i.e. ##h^2## is so much nicer than ##(\Delta y)^2## or having to mess with ##v_{fy}## and ##v_{fx}## and so on.)

    It is also useful to start using latex to write out equations ... the soner you start the aaseir it will be leter when you have to. For the stuff you are doing, it is pretty intuitive. That last equation, for eg, was just v_y=v\sin\theta

    Some things to check:
    by "flying distance" do they mean "the horizontal distance flown" ... it could mean "the length of the path taken through the air".
  17. Apr 20, 2013 #16
    Ok i'm getting confused when trying to solve for time t.

    Δy =-16, so

    Vosin(∅)t + (1/2)(g)t^(2) +16

    but when solving for t how do i know which one to use resulting from the quadratic equation?

    if i use (-Vosin(∅)/g + (Vosin(∅))^(2) -4(1/2)(g)(16))^(1/2)/g i get the max angle at 45,

    but when i use (-Vosin(∅)/g + (Vosin(∅))^(2) -4(1/2)(g)(16))^(1/2)/g and compute angle 45 compared to 46 i get a higher result for 46 than 45.

    Thanks for the advice.
  18. Apr 20, 2013 #17

    Simon Bridge

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    For your notation with "up" being positive, then the acceleration of gravity is negative.
    When you are solving for t from the quadratic equation, one of them will be clearly nonsense.
    You could keep the ##\pm## to the end if you like, this pick the distance that is physically possible.
  19. Apr 21, 2013 #18
    How can i calculate the distance though, without already assuming the angle's value, which also leads me to the question of how can i tell which t is nonsense, without knowing Vo and the angle theta?
  20. Apr 21, 2013 #19

    Simon Bridge

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    Without the initial speed and angle you cannot do this no. However, you worked out what these were in post #6 and post #8.
  21. Apr 21, 2013 #20
    That was for the case where the angle was 30 i.e the angle was given.
    Last edited: Apr 21, 2013
  22. Apr 21, 2013 #21

    Simon Bridge

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    Well the initial speed is the same ... you will leave the angle as a variable.
    You know what range of values it should have.

    You want to end up with an equation which has a distance as a function of angle.
  23. Apr 21, 2013 #22
    ah i see, i was thinking that if i change the angle then h changes. Thanks for clearing that up Simon.
  24. Apr 25, 2013 #23
    mmm still getting confused with this question. So i have x as a function of theta, but i'm getting 44 degrees to give me higher distance then 45. Am i meant to take the derivative of this function and set it to zero then solving for theta?
  25. Apr 25, 2013 #24
    I used wolfram alpha to graph the x v theta graph and its giving me the maximum at 41.79. how do i find this value analytically though, thanks!
  26. Apr 25, 2013 #25

    Simon Bridge

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    If the function has a maximum anywhere, then the slope of the function will be zero at that maximum. You exploit that property to find the maximum.

    Notice that 45deg gives the maximum distance when the final height is the same as the initial height. if T is the time to return to the initial height, then vy(T)=-vsub]y[/sub](0), while vx(T)=vx(0).

    But, in this case, after reaching the initial height again, there is still some more distance to fall in the y direction. The time to fall the rest of the way is fixed. Therefore the extra horizontal distance travelled is determined by the horizontal velocity alone. The bigger that is the farther the distance travelled - and the shallower the initial angle. So it is reasonable to expect that the angle for maximum distance is a bit less than 45deg.

    Notice that if you already know how the distance to the point where the height is the same as the initial height relates to angle, then you only have a simple step to go.
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