Suppose the roller coaster in Fig. 6-41 (h1 = 39 m, h2 = 12 m, h3 = 20) passes point 1 with a speed of 1.70?

m/s. If the average force of friction is equal to one fourth of its weight, with what speed will it reach point 2? The distance traveled is 50.0 m.

m/s

Suppose the roller coaster in Fig. 6-41 (h1 = 39 m, h2 = 12 m, h3 = 20) passes point 1 with a speed of 1.70?
(1/2)mvi^2 + mg?h - (1/4)mgs = (1/2)mvo^2

vo^2 = vi^2 + g(2?h - s/2)

vo^2 = 1.7^2 + 9.80665(2(27) - 50/2)

vo^2 = 2.89 + 9.80665(29)

vo ≈ 16.94942 m/s ≈ 16.9 m/s