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?
- 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
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