1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Answer check: A ball on a spring launched into the air. find speed/height

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    aXvTc.png

    2. Relevant equations

    Since energy is conserved, MEinital = MEfinal

    3. The attempt at a solution
    1/2 mvi2 + mgyi + 1/2kxi2 = 1/2 mvf2 + mgyf + 1/2kxf2

    After finding what equals 0, im left with:
    1/2kxi2 = 1/2mvf2 + mgyf
    v=5.97m/s

    I found yf by: mgyf = 1/2kxi2
    yf = 3.63m


    Just checking to see if these answers are correct. THank you for the help!!!!
     
  2. jcsd
  3. Nov 11, 2012 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I get a different result. Please post more details of your calculation.
     
  4. Nov 12, 2012 #3
    First I found yf, which is the answer to part b.
    Since energy is being conserved, I came up with PEf = KEi. This seems weird because usually there is no potential energy at the final position but it was the only thing I could come up with.

    mgyf = 1/2kxi2
    y= (1/2kxi2) / g
    y=3.63m

    Then,

    1/2kxi2 = 1/2mvf2 + mgyf

    sqrt((kxi2 - mgyf) / m) = v
    v = 5.97m/s
     
    Last edited: Nov 12, 2012
  5. Nov 12, 2012 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Compare those two equations. Shouldn't you get vf=0? ...
    .. but you didn't because you dropped a factor of 2.
    For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.
     
  6. Nov 12, 2012 #5
    oh ok. i understand. the fastest the ball will be going is right after its released from the spring, not at its highest position in the air.

    So its,
    1/2mvi2 + 1/2kxi2 = mgyf
    vi = sqrt( 2gyf - kxi2 )

    But now I'm stuck not knowing what it's final position is.
     
  7. Nov 12, 2012 #6
    Using yfinal = 3.63m. I got the velocity to be 7m/s
     
  8. Nov 12, 2012 #7
    I am wrong. The inal position is .15m because that's how far the spring is compressed an that's how far the ball moves before its released from the spring

    V= 8.3m/s for part a
    Y= 3.6m for part b


    Thanks for the help!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Answer check: A ball on a spring launched into the air. find speed/height
Loading...