Elastic Potential Energy Question

[wmrunner24]

Homework Statement

A pinball machine launches a 90g ball with a spring driven plunger. The game board is inclined at an angle of 6$$\circ$$ above the horizontal.

Assume the plunger's mass and frictional effects are negligible. The acceleration of gravity is 9.8m/s².

Find the force constant k of the spring that will give the ball a speed of 101cm/s when the plunger is released from rest with the spring compressed 4 cm from its relaxed position.
Answer in units of N/m.

Homework Equations

Not too sure, but...
F_g=mgsin$$\Theta$$
F=-kx
$$\sum$$F=ma
And since I see something about speed in there, I assume we'll need a kinematic equation as well? From what I can tell it looks like:
V_f²=V_i²+2ax

The Attempt at a Solution

To start, everything needs converted from cm to m. Next...I'm guessing what needs done is I need to find the acceleration necessary to get the ball from rest to 1.01m/s within 0.04m. Then once that's been found, would I calculate the net force and solve for k in that equation?

 

[kuruman]

Try using conservation of mechanical energy.

 

[wmrunner24]

Ok. So...

$$\frac{1}{2}$$kx²=mgh+$$\frac{1}{2}$$mv²?

 

[kuruman]

Correct. You need to do some trig to find h.

 

[wmrunner24]

Right. Figured. Now, do I need to account for the incline with the force of gravity, or is it good at 9.8?

 

[kuruman]

Right. Figured. Now, do I need to account for the incline with the force of gravity, or is it good at 9.8?

I am not sure what you mean. In mgh, the symbol "h" stands for "change in height". What is the change in height of the ball in this case?

 

[wmrunner24]

I get how to do height. My above question was about gravity. The force of gravity on an incline is mgsin$$\Theta$$. I was wondering if that was necessary here because height is just straight up and down, like gravity.

Oh, and height would be x*sin$$\Theta$$.

 

[kuruman]

It is not necessary. The work done by gravity when the ball moves up the incline by distance d is W_g=-mgdsinθ = -mgh which the negative of the potential energy change. So when you put in the potential energy change as mgh, you are taking care of the effects of gravity.

 

[wmrunner24]

hmm...ok that makes sense. So I should get 61.9903N/m, right? The equation I got was:

$$\frac{m}{x^2}$$(2gxsin$$\Theta$$+V_f²)

 

[kuruman]

I didn't put in the numbers, but your expression looks right.

 

[wmrunner24]

The homework site accepted it as correct. Thank you very much.