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Work-kinetic energy theorem - model rocket velocity/height

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data

    A student experimenting with model rockets measures the speed of a vertically-launched rocket to be 18.0 m/s when it is 75.0 m above the ground on the way up. The rocket engine fires from when the rocket is at ground level to when it is 8.75 m above the ground. If the rocket has a mass of 0.893 kg, use the work-kinetic energy theorem to determine:

    (a) The speed of the rocket when it was 8.75 m above the ground
    (b) The maximum height attained by the rocket

    (make sure you know what the work-kinetic energy theorem is before you start your solution; it does not, for example, use gravitational potential energy.)

    2. Relevant equations

    1. ΔK = ½m(V2 - V02)
    2. Wnet = ΔK
    3. Wgravity = mgh
    4. V = V0 - gt2
    5. ymax = y0 + V0t - ½gt2

    3. The attempt at a solution

    a. My first instinct was to calculate ΔK for the entire flight, which I found to be about 145 J. Though I am not entirely sure this is true because the engine stopped firing at 8.75 m. What pushed me to believe this was the ΔK ,and therefore the Wnet, of the flight up to y = 75.0 m was a quote from my book which reads, " the work-energy theorem is valid even without the assumption that the net force is constant." Unfortunately, I cannot say with certainty that this is even relevant to the problem. If anyone could clarify this I would be very thankful.

    b. That seemed to be a dead end, so I calculated the work done by gravity at 8.75 m and assumed this was the net-work from y = 0 m to y = 8.75 m. But after re-reading the problem I found that to be foolish, and so I was back at square one.

    c. I became a bit distressed and decided to hit point a again, I figured this:

    If Wnet = 145 J at y = 75.0 m , then the engine must have done upward work from y = 0 m to y = 8.75 m such that,
    Wengine - Wgravity (which is about 657 J at y = 75.0 m) = 145 J.

    So Wengine must be,

    Wnet + Wgravity = 802 J.

    If this were true, then:

    Wengine = ΔK0→8.75m

    Wengine = ½m(V2 - 0) → √(2Wengine)/m = V8.75m → V8.75m = 42.4 m/s

    d. If this were all true, then the maximum height would be achieved when,

    Wgravity = Wengine (a bit irrelevant but I'd like to know if this is true)

    and I could use equation 4 above to solve for the time of the flight from y = 8.75 m to ymax, which I could then put into equation 5 where y0 = 8.75 m, and V0 = 42.4 m/s

    I hope at least one of the conclusions I drew will be of actual use (though I'm doubtful). Thank you in advance for pointing me in the right direction!
     
  2. jcsd
  3. Nov 3, 2015 #2

    andrevdh

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    Homework Helper

    Since you can solve an equation if it has only one unknown maybe you should apply the theorem between one point where the speed is known and one point where the speed is required.
     
    Last edited: Nov 3, 2015
  4. Nov 3, 2015 #3
    So then I should calculate the work done by gravity on the rocket between 8.75 m and 75.0 m, and then set that equal ΔK to solve for the velocity at 8.75 m?
     
  5. Nov 3, 2015 #4

    JBA

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    You are on the right track.
     
  6. Nov 4, 2015 #5

    andrevdh

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    Homework Helper

    That is how I thought one could solve for the speed of the rocket just as it stopped firing, yes.
     
  7. Nov 4, 2015 #6

    JBA

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    You are correct. The maximum velocity is achieved at the height that the rocket stopped firing.
     
  8. Nov 4, 2015 #7
    I've figured it out! Thanks for the input.
     
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