Question

A 57.4 g pinball starts at the top of a frictionless pinball table inclined at 50.0°. If the ball is released and the spring at the bottom is compressed 0.0800 m and has a spring constant of k = 21.5 N/m, how far back up the pinball table will the pinball travel before rolling back down again?

Answer 1

let the height of oncline = h

1) when all released from top

KEi + PEi = KEf+ PEf

0 + mgh = 0 + 1/2 k x²

57.4 x 10^-3 xgh = 1/2 x 21.5x (0.08)²

h = 0.122 m

as the inclined is friction less so when the spring will be normal again the stored energy will convert in PE

so

now ( till the point the pinball travel before rolling back down again at that point againt the v = 0)

so h = same as befor

h = 0.122 m

so traveled distance = d = h/sin50 = 0.122/sin50 = 0.16 m