Solving Pinball Machine Spring Force Problem

[wallercp]

Homework Statement

The ball launcher in a pinball machine has a spring that has a force constant of 1.2 N/cm. The surface on which the ball moves is inclined at 10° with respect to the horizontal. If the spring is initially compressed 5 cm, find the launching speed of a .1 kg ball when the plunger is released. Friction and the mass of the plunger are negligible.

Homework Equations

Ue (Elastic potential) = (1/2)kx^2
K (Kinetic) = (1/2)mv^2
Ug (Gravitational potential) = mgh

The Attempt at a Solution

k = 1.2 N/ cm = .012 N / m
x = 5 cm = .05 m
m = .1 kg
g = 9.8 m/s

Ue = K + Ug

.5kx^2 = .5mv^2 + mgh

.5(.012)(.05^2) = .5(.1)v^2 + .1(9.8)(.05sin(10°))

0.000015 = .05v^2 + .00851

At this point, my v^2 will equal a negative number, which makes no sense at all. I'm stuck >.<

The problem is that the left side of the equation gets exponentially smaller while the right hand side... doesn't.

Edit: Foolish calculator mistake. I hate how I find this out right after I posted it (I checked my work for 1/2 hr before posting).

1.2 N / cm = 120 N / m, not .012. I hit * instead of / on my calc.

 

[Ja4Coltrane]

check your conversion from n/cm to n/m

 

[conquertheworld5]

I think it's just your conversion for the k value; you have to multiply
1.2N/cm*100cm/1m = 120N/m

my bad... didnt see that someone already replied.