Simple harmonic motion pinball problem

[heather32283]

I have been looking at this problem forever. It is probably easy, but I am making it harder. States: In preparation for shooting a ball in a pinball machine, a spring (k=675 N/m) is compressed by 0.0650m relative to its unstrained length. The ball ( m=0.0585kg) is at rest against the spring at point A. When the spring is released, the ball slides (w/o rolling) to point B, which is 0.300m higher then point A. How fast is the ball moving in B?

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

 

[quasar987]

Do it using conservation of energy.

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.

 

[heather32283]

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

 

[quasar987]

Yeah, you rock! Then use the fact that $E_f = E_0$ to recover $v_f$.

 

[heather32283]

what do u mean to recover vf

 

[heather32283]

so 1/2(.058kg)(0 m/s) + 1/2(.058kg)(9.80 m/s2)(h?) + 1/2(675 N/m)(x?)2
so xo=.0650 and then xf is .300 m higher so xf=.3650m right

 

[quasar987]

lol, what? Be consistent with your notation. You got 2 values of xf there.
What you got to do is set Ef = Eo, plug all the values, solve for vf.

 

[heather32283]

no the problem says that point B is .300m higher then a so I thinmk that would mean that xf would = .3650m bc u would add the beginning xo value plus the .300m value. Also I don't know what h would be and xfor the equation

 

[heather32283]

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

 

[quasar987]

h is 0.300 and the Xo of the spring's pot.energy is 0.0650m.

The problem is 2 dimensional. When they say point B is 0.300m higher, it means higher in the sense of 'h'.

 

[quasar987]

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

No you're missing the gravitationa potential energy now. You had it right in your earlier post.

 

[heather32283]

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

 

[quasar987]

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

 

[heather32283]

opps had it right on paper but put it in the wrong area on here. So would xf=.650 +.300, that is the only variable I am trying to find, other then the final speed

 

[quasar987]

Xf is 0. It represent the length by which the spring is stretched! As soon as the ball leaves the spring x is 0.

 

[heather32283]

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

 

[quasar987]

ho = 0

xo = 0.0650

 

[heather32283]

so we plug in the numbers and solve for vf right

 

[quasar987]

right...

 

[heather32283]

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

 

[heather32283]

does 27.4 m/s sound right