- #1

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you want to find the vf (velocity final), I think so I set up the equation as 1/2vf2 + 1/2 kxf2= 1/2 kxo2 (o=initial, f=final)

vf=sqrt km(xo2-xf2) which I got an outrageous number for. Please help I am soo lost

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- Thread starter heather32283
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In summary, the conversation discusses a problem involving a ball being shot from a spring in a pinball machine. The spring has a constant of 675 N/m and is compressed by 0.0650m. The ball has a mass of 0.0585kg and is initially at rest against the spring at point A. When released, the ball slides to point B, which is 0.300m higher than point A. The conversation then discusses how to calculate the speed of the ball at point B, using conservation of energy. It is determined that the final velocity can be found by setting the final energy equal to the initial energy, and solving for the final velocity. After some confusion and calculation, the final velocity is determined

- #1

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you want to find the vf (velocity final), I think so I set up the equation as 1/2vf2 + 1/2 kxf2= 1/2 kxo2 (o=initial, f=final)

vf=sqrt km(xo2-xf2) which I got an outrageous number for. Please help I am soo lost

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- #2

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1) Calculate the potential energy stored in the spring when strechted. Once released, all that energy becomes kinetic.

2) Calculate what energy the ball has converted in gravitational potential once it reaches B.

3) The remains of the energy is still kinetic. Find the velocity associated to that amount of kinetic energy.

- #3

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So you would set up the equation as E= 1/2MV2 + mgh + 1/2kx2 right

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Yeah, you rock! Then use the fact that [itex]E_f = E_0[/itex] to recover [itex]v_f[/itex].

- #5

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what do u mean to recover vf

- #6

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so xo=.0650 and then xf is .300 m higher so xf=.3650m right

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What you got to do is set Ef = Eo, plug all the values, solve for vf.

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- #9

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so we would use 1/2mvf2 + 1/2 kxf2= 1/2mvo2 +1/2 kvo2 from the equation Ef =Eo

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The problem is 2 dimensional. When they say point B is 0.300m

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No you're missing the gravitationa potential energy now. You had it right in your earlier post.heather32283 said:so we would use 1/2mvf2 + 1/2 kxf2= 1/2mvo2 +1/2 kvo2 from the equation Ef =Eo

- #12

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/2mvf2 + mgho +1/2 kxf2= 1/2mvo2 + mghf +1/2 kvo2 right

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heather32283 said:1/2mvf2 + mgho +1/2 kxf2= 1/2mvo2 + mghf +1/2 kxo2 right

Ef = Eo means (all final values) = (all initial values). Not a mix of both. Keep it to

1/2mvf2 + mghf +1/2 kxf2= 1/2mvo2 + mgho +1/2 kxo2

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- #16

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so hf=.300, ho= .0650 ??

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ho = 0

xo = 0.0650

xo = 0.0650

- #18

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so we plug in the numbers and solve for vf right

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right...

- #20

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that will then give us the speed at B, god you have been such a help thank you

- #21

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does 27.4 m/s sound right

Simple harmonic motion is a type of periodic motion in which a system moves back and forth around a central equilibrium point, with a restoring force proportional to the displacement from the equilibrium position.

In a pinball machine, the ball moves back and forth in a pendulum-like motion as it bounces off the pins and bumpers. This motion is caused by the force of gravity pulling the ball down and the elastic force of the bumpers pushing it back up, creating a simple harmonic motion.

The factors that affect simple harmonic motion in a pinball machine include the gravitational force, the elasticity of the bumpers and pins, and the initial velocity and direction of the ball.

The equations of motion, such as the equation for position, velocity, and acceleration, can be used to calculate the position, velocity, and acceleration of the ball at any given time during its motion. This information can then be used to analyze the behavior of the ball in the pinball machine.

Yes, the simple harmonic motion of a pinball machine can be affected by external factors such as air resistance, friction, and the shape and layout of the machine. These factors can alter the motion of the ball and make it deviate from a perfect simple harmonic motion.

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