- #1

desichick07

- 18

- 0

## Homework Statement

The ball launcher in a pinball machine has a spring that has a constant of 1.2 N/cm. The surface on which the ball moves is inclined 10 degrees with respect to the horizontal. If the spring is initially compressed 5.0 cm, find the launching speed of a 100g ball when the plunger is released.

## Homework Equations

W = .5kx^2

mgh + .5kx^2 + .5mv^2 = mghf + .5kxf^2 + .5mvf^2

## The Attempt at a Solution

first i converted the spring constant

1.2 N/cm * 100cm/m = 120 N/m

next i found PE(spring)

.5kx^2 = .5(120N/m)(.05m)^2 = .15 J

then i used the mechanical energy eq...i canceled the unnecessary terms and got:

.5kx^2 + .5mv^2 = mgh(f)

and when i solve for V(initial) i get:

((mgh(f) - .5kx^2 )/(.5m))^.5

but then i don't know the distance which the ball travels so i can't use this equation

and also i know the 10 degrees has to factor in somewhere, but where? is it part of the potential energy of the string? or does it affect the initial velocity? or both?