1. Not finding help here? Sign up for a free 30min 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!

Spring constant and velocity

  1. Feb 25, 2015 #1

    Yam

    User Avatar

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

    The force constant of a spring is 600 N/m and the un-stretched length is 0.72 m. A 3.2-kg block is suspended from the spring. An external force slowly pulls the block down, until the spring has been stretched to a length of 0.86 m. The external force is then removed, and the block rises. In this situation, when the spring has contracted to a length of 0.72m , the upward velocity of the block is = ???

    2. Relevant equations
    Conservation of Energy
    Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

    3. The attempt at a solution

    0 + 0.5(600)(0.14)^2 = 0.5 x 3.2 x v^2 + 0
    v = 1.92m/s?

    This is incorrect. The four options given are
    a) 1.0 m s b) 5.1m s c) 7.4 m s d) 9.1 m s e) 9.7 m s


    Do I need to account for gravitational potential energy in the equation?
     
    Last edited: Feb 25, 2015
  2. jcsd
  3. Feb 25, 2015 #2

    Merlin3189

    User Avatar
    Gold Member

    I think you may have mis-written your "Relevant Equation" , then when you substitute the numbers you got a bit mixed up.

    Perhaps you could state each potential energy separately.

    Edit: -oops, sorry I didn't read your final comment. Can you link to the original Q please?
     
  4. Feb 25, 2015 #3

    Yam

    User Avatar

    Hi, sorry for the mistake.
     
  5. Feb 25, 2015 #4

    Merlin3189

    User Avatar
    Gold Member

    Are you saying,
    PE1 +KE1 = PE2 +KE2
    0 + 0.5(600)(0.14)^2 = 0.5 x 3.2 x v^2 + 0

    If so, a) when or where is "Initial" and when is "Final" ?
    b) looking at your formula, one of your expressions is for KE1 and the other for PE2, but both look like 0.5 m v2
    (though, what mass is 600?)

    Perhaps you should state clearly where you are for each expression before you put it into an equation.

    You might also think about what sorts of PE exist.
     
  6. Feb 25, 2015 #5

    Yam

    User Avatar

    PE1 = Elastic potential energy = 0.5 * k * x^2 =0.5(600)(0.14)^2
    KE1 = 0

    PE1 = gravitational potential energy = mass * g *height = (3.2)*(9.8)*(0.14)
    KE2 = 0.5*m*v^2 = 0.5 x 3.2 x v^2

    So,

    0.5(600)(0.14)^2 = (3.2)*(9.8)*(0.14) + 0.5 x 3.2 x v^2

    v = 0.96m/s?
     
  7. Feb 25, 2015 #6

    Merlin3189

    User Avatar
    Gold Member

    That looks ok now.
    Presumably they've rounded the answer to 1dp.
     
  8. Feb 25, 2015 #7

    Yam

    User Avatar

    Thank you for your help! Stating clearly for each expression before putting into an equation really helps!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted