What is the Velocity of a Shank with Attached Spring Striking a Button?

[doombanana]

Homework Statement

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.

Homework Equations

$E_1=E_2$
$PE_{spring} = \frac{1}{2}k(x_1-x_0)^2$
$PE_{gravity} = mgh$
$KE = \frac{1}{2}mv^2$

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:

$PE_{spring,1} + PE_{gravity,1} = PE_{spring,2} + PE_{gravity,2} + KE_2$

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

I get 4.27 ft/s.

The correct answer should be 3.43 ft/s. I'm just not sure where I went wrong

 

[TSny]

$PE_{spring} = \frac{1}{2}k(x_1-x_0)^2$

Did you find the value of $x_0$?

10 lb/in = 120 lb/ft

$PE_{spring,1} + PE_{gravity,1} = PE_{spring,2} + PE_{gravity,2} + KE_2$

$\frac{1}{2}(1.2)(.125)^2 + (5)(.125) = \frac{1}{2}(1.2)(-.0208)^2 - (5)(.0208) + \frac{1}{2}(\frac{5}{32.2})v^2$

Did you mean to let k = 120 lb/ft here?

 

[doombanana]

Did you find the value of $x_0$?

Ah, that's what it was. I forgot that the spring is compressed at the equilibrium point.

Thank you again!