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Conservation of energy with a rocket disagreement

  1. Mar 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1Kg rocket is fired off. The engine provides a thrust of 18 Newtons for 20 meters. What is the maximum height achieved by the rocket? Assume no loss of mass and no friction. Gravity=9.8m/s^2

    2. Relevant equations
    Work=Force x distance
    Force=Mass x acceleration(or gravity)
    Potential gravitational energy(PE)=Mass x Gravity x Height
    Kinetic energy(KE)=(1/2) x Mass x Velocity^2


    I have a disagreement with my teacher on how to correctly solve this type of problem. Both of us agree that I need to find the work done by the engine of the rocket during the first 20 meters of flight first (W= 18N*20m=360 joules). After this is where my teacher and I begin to differ. My teacher says that I need to then set the work done by the engine = to the maximum PE of the rocket and solve for the height(W=PE so maximum height= W of engine / (Mass x Gravity) = 360/(1*9.8)= 36.7 joules). I believe that in this solution energy is not conserved.

    From what I understand, since the rocket only stops accelerating at 20 meters, the work of the engine up to this point should be set = to its KE. This gives you a KE 360 joules. The rockets PE, if measured from the highest point of its flight down to 20 meters, is 360 joules. This value does not account for the 20 meters before the rocket begins to be decelerate from gravity when it is still accelerating therefore you must find the PE of the rocket at 20 meters (9.8m/s^2 * 20m * 1Kg = 196 joules) and add that onto the KE at 20 meters in order to have the maximum PE for the rocket (556 joules) since it will convert that KE into PE in a 1:1 rate. Using this value for PE you get a maximum of 56.7 meters.

    Which of us is right with this? Is my logic sound in this matter?
     
  2. jcsd
  3. Mar 4, 2015 #2

    Bystander

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    What's the net force acting on the rocket?
     
  4. Mar 4, 2015 #3
    The 18 newtons is assumed to have already accounted for gravity in my textbook so is the net force acting on the rocket during the first 20 meters.
     
  5. Mar 4, 2015 #4

    TSny

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    It appears to boil down to the interpretation of the "thrust". If the thrust is the force that the rocket receives from the engine alone, then your teacher is correct. If the thrust is the net force on the rocket (engine + gravity) then you are correct.

    I believe the usual interpretation of thrust is the force of the engine only. Thus, if the engine were adjusted so that the rocket hovers at rest above the earth, then I think most people would say the magnitude of the thrust of the rocket equals the magnitude of the force of gravity rather than saying the thrust is zero.
     
  6. Mar 4, 2015 #5
    Thank you for the help.
     
  7. Mar 5, 2015 #6

    kuruman

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    It simply is not true that "since the rocket only stops accelerating at 20 meters, the work of the engine up to this point should be set = to its KE." Gravity also does work over the 20 m distance.

    The best way to look at this is to forget mechanical energy conservation and use the (more general) work-kinetic energy theorem, WNet = ΔK. Here WNet = WEngine+WGravity = F*d - mghmax and ΔK = 0.

    This results in the teacher's solution.
     
  8. Mar 5, 2015 #7

    TSny

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    Right, the rocket doesn't stop accelerating when the engine shuts off.

    According to post #3, Snakeish says that the the 18 N "thrust" is the net force on the rocket (engine force + gravity force). That's an odd interpretation of "thrust", but if that's what the 18 N is meant to denote, then the work done by the 18 N force would equal the KE of the rocket at the point the rocket shuts off. This would lead to Snakeish's answer.

    But I'm with you, kuruman. The teacher's answer corresponds to the more natural interpretation of the 18 N force.
     
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