Question

Roller coaster problem is set up as such...Point A is 10m above ground, distance to Point B is 40m and Point B touches the ground, distance to Point C is 20m and Point C is 8m above ground. A 500kg roller coaster starts from rest at Point A and begins to coast down the track. If the retarding friction is 40N, how fast will roller coaster be going at Point B and Point C?

Answer 1

a)
use:
potential energy at A = kinetic energy at B + energy lost against friction
m*g*h = 0.5*m*v^2 + f*d
500*9.8*10 = 0.5*500*v^2 + 40*40
49000 = 250*v^2 +1600
v = 13.77 m/s
Answer: AT point B, speed is 13.77 m/s

b)
use:
potential energy at A = kinetic energy at c +potential energy at c + energy lost against friction
m*g*h = 0.5*m*v^2 + m*g*h' + f*d
500*9.8*10 = 0.5*500*v^2 + 500*9.8*8 + 40*(40+20)
49000 = 250*v^2 + 39200 + 2400
v = 5.44 m/s
Answer: AT point C, speed is 5.44 m/s