A frictionless roller coaster is designed with an initial drop of 135m, and the

Question

A frictionless roller coaster is designed with an initial drop of 135m, and the coaster begins at 11m/s. Find:
a) Velocity at the bottom of the coaster.
B) Velocity at point 25m from the ground.
c) Can the coaster make it up a loop the loop that has a radius of 72m?(the bottom of the loop is at the ground level)
d) If at the bottom of the roller coaster there is a cushioning spring intended to stop the coaster with a spring constant 10^4N/m, calculate the compression of the spring when the coaster gets in contact with it. The mass of the roller coaster Car is m=100kg.

Explanation / Answer

a)

h = 135 m

Vt = speed at top = 11 m/s

Vb = speed at bottom

using conservation of energy

Kinetic energy at top + Potential energy at top = Kinetic energy at bottom

(0.5) m Vt2 + mgh = (0.5) m Vb2

Vb2 = Vt2 + 2gh

Vb2 = 112 + 2(9.8)(135)

Vb = 52.6 m/s

b)

h = 135 m

Vt = speed at top = 11 m/s

V = speed at 25 m mark

h' = 25 m

using conservation of energy

Kinetic energy at top + Potential energy at top = Kinetic energy at 25 m mark + PE at 25 m mark

(0.5) m Vt2 + mgh = (0.5) m V2 + mgh'

(0.5) Vt2 + gh = (0.5) V2 + gh'

(0.5) (11)2 + (9.8)(135) = (0.5) V2 + (9.8)(25)

V = 47.72 m/s

c)

r = radius of loop = 72

for looping the loop , speed required at bottom = sqrt(5gr) = sqrt (5 x 9.8 x 72) = 59.4 m/s

since Vb = 52.6 , hence coaster will make it up

d)

using conservation of energy

Kinetic energy at top + Potential energy at top = spring potential energy

(0.5) m Vt2 + mgh = (0.5) k X2

(0.5) (100) (11)2 + 100 (9.8) (135) = (0.5) (10000) X2

X = 5.3 m

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