A roller coaster of mass m =540 kg is to be launched by a spring which is compre

Question

A roller coaster of mass m=540 kg is to be launched by a spring which is compressed a distance d=7.0 m. Upon being released the roller coaster is accelerated along a frictionless track by the spring, and then goes up a hill to a height h=2.5 m, where the track has a radius of curvature of R=5.6 m as shown. The riders are to fell weightless as they go over the hill. What magnitude of force does the compressed spring exert on the roller coaster while it is waiting to be launched?

The answers was 8000N can somone please explain how?

Explanation / Answer

given,

m = 540 kg

d = 7 m

R = 5.6

if riders have to feel weightless then

mg = mv^2/R

g = v^2/R

9.8 = v^2 / 5.6

v = 7.41 m/s

by conservation of energy

initial energy = final energy

0.5 * k * x^2 = 0.5 * mv^2 + mg(h + 2R)

0.5 * k * 7^2 = 0.5 * 540 * 7.41^2 + 540 * 9.8 * (2.5 + 2 * 5.6)

k = 3564.31 N/m

force = kx

force = 3564.31 * 7

force spring will exert = 24950.17 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.