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# A 44.2 g pinball starts at the top of a frictionless pinball table inclined at 45.0°?

A 44.2 g pinball starts at the top of a frictionless pinball table inclined at 45.0°. If the ball is released and the spring at the bottom is compressed 0.0500 m and has a spring constant of k = 24.5 N/m, how far back up the pinball table will the pinball travel before rolling back down again?

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- civil_av8rLv 73 years agoFavorite Answer
Change in energy = 0

Change in GPE + Change in SPE = 0

M*g*(hf - hi) + 1/2*k*(xf^2 - xi^2) = 0

Eliminate some variables

Hf-hi = h

Xf = 0

So,

M*g*h - 1/2*k*xi^2 = 0

Solve for h

h = (k*xi^2)/(2*m*g)

Finally, the height is a component of length.

H = L*sin (theta)

Plug in and solve for L

L = (k*xi^2)/(2*m*g*sin (theta))

Now plug in numbers

L = 0.100 m

Thats what I got

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