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200kg Astronaut, 2.0 m/s, 100.0 N rocket ?

  1. Mar 9, 2008 #1
    [SOLVED] 200kg Astronaut, 2.0 m/s, 100.0 N rocket ?

    A 200.0 kg astronaut and equipment move with a velocity of 2.00 m/s toward an orbiting spacecraft. How long will the astronaut need to fire a 100.0 N rocket backpack to stop the motion relative to the spacecraft? Give your answer in s.


    2. Relevant equations
    F=ma
    W=mg
    m=w/g
    a=F/m


    3. The attempt at a solution
    I know I am missing something so simple. I have spent too long trying to solve this problem and have given up.

    F=(200kg)(2 m/s)=400 kg*m/s=
    F=400N/s

    I don't know what to do with the 100N for the rocket pack, I'm getting lost in all the different conversions.

    Please Help
     
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2
    Why don't you start by applying 100N for one second, and see what the final speed is?

    There is definitely a way to apply the equations, but understand it intuitively and it'll make more sense.

    Also, in your "F=(200kg)(2 m/s)=400 kg*m/s^2", you have an extra 1/s on the right side.
     
  4. Mar 9, 2008 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Look again at the units. That should be [itex]200\,\text{kg} 2\,\text{m}/\text{s} = 400 \,\text{kg-m}/\text{s}[/itex]. The product of velocity and mass is momentum, not force. An alternative expression for kg-m/s is Newton-seconds. That 400 kg-m/sec is the same as 400 Newton-seconds, and this form should tell you exactly how long the rocket pack needs to be fired.
     
  5. Mar 9, 2008 #4
    So if the Astronaut has a Force of 400 Newton-seconds and he fires a 100 N rocket then the answer will be 4 s?

    If that is the correct answer I'm going to be pissed that is the solution I came up with when I first began working this problem. But, thought that it was to easy and couldn't be right.
     
  6. Mar 9, 2008 #5
    Pretty much. For each 100N force applied for 1 second, the decrease in velocity is 0.5m/s.

    In 4 seconds, -2.0 m/s decrease in velocity.

    Enjoy.
     
    Last edited: Mar 9, 2008
  7. Mar 9, 2008 #6
    Thanks
     
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