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Ball falling on Spring.

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data
    A sphere of 4kg falls from a height of 70cm and lands on a spring of constant 2 *10² N/m.
    Whats the max velocity the sphere reachs?
    g=10m/s²


    2. Relevant equations
    Ep=m*G*h
    Ek=mv²/2
    Eel=k*x²/2
    Vmax=sqrt(2g*h) (internet source idk if this is right)


    3. The attempt at a solution
    I tried this Ep = Eel + epx
    Ep=4*0,7*10
    28=kx²/2 + mg*x
    x=0,367 m
    them i did 0,7 + 0,367 for the height
    and it gave me Sqrt(2g*h)=v
    V=4,62 m/s
    not the answer, i couldnt get my head around the energies. I also tried Ep=Eel but no sucess
     
  2. jcsd
  3. Dec 5, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    These are OK.

    This only applies to a body falling freely from a height h. Don't need it.


    The total energy is (using your notation): Ep + Ek + Eel
    That total energy is conserved.

    Hint: At what point does the mass stop accelerating? (After that point, it starts slowing down.)
     
  4. Dec 5, 2013 #3
    Sorry, power went out on my house.
    So it stops acelerating when it touches the spring.
    But i still need the energy of the spring right because it touches it?
    If i try ep+ek + Eel i get 28 -2v² = 200x²/2
    now to find the x
    i can use the weight of the ball F=m*a = k *x , i get 40=200*x => x=0,2
    them Eel = k*x²/2 = 4 !
    Eel+ Ep = Ec Ec = 32
    32= 4 * v² /2
    16 = v²
    v=4 m/s Yeaah lol solved while writing this thank you veryy much ! (or maybe i did something wrong and got to the right result ?)
     
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