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!

Shank with attached spring

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data

    The shank of a 5-lb vertical plunger is .25 inches above a button when resting in equilibrium against the spring of stiffness k = 10 lb/in. The upper end of the spring is welded to the plunger, and the lower end is welded to the base plate. If the plunger is lifted 1.5 inches above its equilibrium position and released from rest, calculate its velocity as it strikes the button A. Friction is negligible.

    2. Relevant equations
    [itex] E_1=E_2[/itex]
    [itex] PE_{spring} = \frac{1}{2}k(x_1-x_0)^2[/itex]
    [itex] PE_{gravity} = mgh[/itex]
    [itex] KE = \frac{1}{2}mv^2[/itex]

    3. The attempt at a solution
    let y=0 at the equilibrium position .25 inches above the button.

    1.5in = .125 ft
    10 lb/in = 120 lb/ft
    .25 in = .0208 ft

    It is released from rest, so KE1=0.

    My energy balance ends up being:

    [itex] PE_{spring,1} + PE_{gravity,1} = PE_{spring,2} + PE_{gravity,2} + KE_2[/itex]

    Plugging in numbers gives:
    [itex] \frac{1}{2}(1.2)(.125)^2 + (5)(.125) = \frac{1}{2}(1.2)(-.0208)^2 - (5)(.0208) + \frac{1}{2}(\frac{5}{32.2})v^2[/itex]

    I get 4.27 ft/s.

    The correct answer should be 3.43 ft/s. I'm just not sure where I went wrong
  2. jcsd
  3. Mar 25, 2013 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Did you find the value of ##x_0##?

    Did you mean to let k = 120 lb/ft here?
  4. Mar 25, 2013 #3
    Ah, that's what it was. I forgot that the spring is compressed at the equilibrium point.

    Thank you again!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted