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Homework Help: Rocket Thrust From a Spring (Work)

  1. Mar 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A 11.7 kg weather rocket generates a thrust of 240 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 560 N/m, is anchored to the ground.

    (a) Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed?

    20.5cm (.205m)

    (b) After the engine is ignited, what is the rocket's speed when the spring has stretched 33.0 cm?


    (c) For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring?

    This is the question I have a problem with.

    2. Relevant equations

    Ug = mgh
    Us = 1/2 k (delta_x)^2
    KE = 1/2mv^2
    W_thrust = F_thrust*distance

    3. The attempt at a solution

    Ok, so for Part (b) I used the following equation to find the velocity:

    W_thrust + U_s(using x_1) = U_s(using x_2) + U_g + KE

    I was able to solve for the v in KE correctly using this equation. So in order to find the velocity of the rocket when the spring is taken out, I take the U_s out of the equation:

    W_thrust = U_g + KE Where the distance here will be x_1 + x_2. (x_compressed + x_stretched = x_total)

    I've done this problem a number of ways.. work, energy, kinematics.. and i keep arriving at the (incorrect) answer v = 3.38m/s. Can anyone help me spot my error? Thanks in advance.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Mar 13, 2008
  2. jcsd
  3. Mar 13, 2008 #2
    I think the answer is 3.79 m/s
    Last edited: Mar 13, 2008
  4. Mar 13, 2008 #3
    Nope, that doesn't work.
    Here's what i have for the energy equation.

    Work_thrust = U_g + KE

    (240N)(.535m) = (11.7kg)(9.81)(.535m) + 1/2(11.7kg)v^2

    With that equation I get 3.38m/s, which is wrong. Maybe I'm using the wrong height? The problem asks for the velocity at the height of the previous question, where x1 was -.205m and x2 was +.33m, so the deltaX would just be .535, right?
  5. Mar 13, 2008 #4
    I think I figured out the problem, one sec....

    Ok, I get 3.79 m/s. Is this correct?

    If it is, then I am pretty sure that I know what you did wrong and why all the other methods you used gave you the same answer.
    Last edited: Mar 13, 2008
  6. Mar 13, 2008 #5
    Nope.. That was my last attempt, the answer turned out to be 3.67. Any idea what we're doing wrong?
  7. Mar 13, 2008 #6
    I thought thrust was a force. Knowing force and mass, we know the rocket's acceleration. Knowing that, we know velocity at any given distance. Where am I going wrong?



    dx/dt=2.6018 m/s
    Last edited: Mar 13, 2008
  8. Mar 14, 2008 #7
    You're forgetting gravity.

    Nothing after dx/dt^2=a makes sense. This is a second order differential
    equation, of wich the solution is x = (1/2) a t^2
  9. Mar 14, 2008 #8
    I think the spring will still push against the rocket, so you can add the energy of
    the compressed spring to the rocket.
  10. Mar 14, 2008 #9
    I just came within 0.262% of that number

    3 components of force going on here:

    1) the rocket's thrust
    2) the spring's push (as kamerling says, the spring is still pushing)
    3) the earth's pull

    acceleration from rocket thrust


    acceleration from spring's push



    the earth's pull


    Now add them all up and you get


    error margin of 0.272%
    Last edited: Mar 14, 2008
  11. Mar 15, 2008 #10
    The acceleration of the spring isn't constant. The acceleration of the spring only works from x= -20.5 up to x=0. You only computed x and v when x>=0, but the initial position of the rocket is at -20.5.

    Because of the non-constant acceleration of the spring it's rather hard to compute x(t) or v(t) and it's easier to compute the kinetic energy of the rocket, wich comes from.
    - The energy of the compressed spring.
    - The force of the rocket engine wich acts over a 0.535 m stretch.
    - The force of gravity wich also acts over a 0.535 m stretch, but in the opposite direction of the velocity, so its contribution is negative.

    If you add these up, and use (kinetic energy) = (1/2)mv^2 you get the right answer.
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