# Spring+incline plane question

1. Nov 21, 2005

### 404

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?

2. Nov 21, 2005

### Staff: Mentor

(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.)

3. Nov 21, 2005

### 404

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

4. Nov 21, 2005

### Staff: Mentor

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

5. Nov 21, 2005

### 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.

6. Nov 21, 2005

### 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?

7. Nov 21, 2005

### 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).

8. Nov 21, 2005

### 404

okay, thanks.