Spring+incline plane question

[404]

The ball launcher in a pinball machine has a spring with a force constant of 1.2N/m. The surface on which the ball moves is inclined 10 degrees with respect to the horizontal. If the spring is initially compressed 5 cm, find the launch speed of a 0.1kg ball when plunger is released. Ignore friction + mass of plunger.

My work so far...

KE(1) + 0.5(k)(x)^2 = Work (gravity) + PE(2)

where 1 is the bottom of the incline plane and 2 is the top. and 0.05m is the distance between the 2 points. (using bottom of incline plane as reference point)

I found the hight at 0.05m up the incline to be 0.0087m and the x component of gravity to be 0.17N

so...
0.5(0.1)V^2 + 0.5(1.2)(0.05^2) = .17*0.05 + (0.1)(9.8)(0.0087)

and I found to be to 0.557m/s, but it's suppose to be 1.68m/s

Where did I go wrong?

 

[Doc Al]

You made several errors:
(1) The initial KE is zero. It starts from rest. The initial energy is spring PE.
(2) The final energy is a mix of gravitational PE plus KE.
(3) If you include gravitational PE as a form of energy, then you don't separately include the work done by gravity. (The gravitational PE is the work done by gravity! To include both is to count it twice.)

 

[404]

but if the initial KE is 0, how do I find the initial velocity that was suppose to be 1.63m/s?

 

[Doc Al]

You are asked to find the launch speed of the ball after it leaves the spring.

 

[mezarashi]

The confusion is coming from how you define "initial" If you define initial as when the spring was still compressed and held in place, the kinetic energy is zero, the spring potential is at $$U = \frac{1}{2}kx^2.$$, which when let go, will convert into the ball's kinetic energy completely.

 

[404]

Well how do you set it up then? I'm more confused then before I posted this now :(

so is it
PE(1) = KE(2) + PE(2)?
where PE(1) = 0.5Kx^2?

 

[mezarashi]

Yes, and more precisely:

PE1 + KE1 = PE2 + KE2

But we know that KE1 is zero, since the ball is at rest when it is in at the spring (which is compressed).

 

[404]

okay, thanks.