Pinball machine problem on energy/power/work

[brnnpink4]

Homework Statement

A pinball machine has a spring that loads up, and shoots a small ball up a shallow incline. The mass of the pinball is .05kg. The incline for the pinball is 15 degrees. Assume no friction exists along the distance over which the spring applies its force. The coefficient between the ball and sloped surface is .15. The speed of the ball at the top of the incline is 3m/s. The spring is compressed .08 m. The length of the incline is .75m. What is the value of the spring constant?

Homework Equations

PE=mgh
KE=.5mv^2
KE+PE=w
Spring constant

The Attempt at a Solution

.05(9.8)h=PE

KE=.5(05)(9)

 

[jackarms]

This is another conservation of energy problem: looks like you have both kinetic energy and potential energy being gained, whereas elastic potential energy from the spring is being lost. You can calculate the potential energy of a spring with this equation:

$U_{e} = \frac{1}{2}k(x_f^{2} - x_i^{2})$, where $x_f$ is the final extension of the spring, and $x_i$ is the initial extension, and $k$ is the spring constant.

 

[brnnpink4]

How would you go about solving xf and xi?

 

[jackarms]

Just think of what they mean: final extension is just that: final extension. After the spring decompresses and launches the ball, it's not compressed anymore, so final extension is zero. Initial extension is also just that. Look at givens in the problem for how much the spring is compressed to start.

 

[HalfLostAndSearching]

3^2

KE+PE= w

1.47+.245= w
w=1.715

KE+PE= w

.5(.05)(9.8)h+.245=1.715

.245+.5(4.9)h=1.715

.5(4.9)h=1.47

4.9h=1.47

h=.3

PE=.05(9.8)(.3)

PE=.147

KE= .5(.05)(9)3^2

KE=1.47

.147+1.47=w

w=1.617

w=1.5(.08)

w=1.617

F=kx

1.617= k(.08)

k= 1.617/.08

k=20.2125

The value of the spring constant for this pinball machine is approximately 20.2125 N/m. This means that for every 1 meter the spring is compressed, it will exert a force of 20.2125 Newtons. This value can also be used to calculate the potential energy and kinetic energy of the ball as it travels up the incline and reaches the top. It is important to note that this value may vary slightly depending on the accuracy of the measurements and assumptions made in this problem.