Question

The Kingda Ka rollercoaster at 6 flags holds the record for the longest drop off with 127m drop.

A. If the coaster starts at the top of the hill with an initial velocity of 2 m/s, assuming no work is done by non-conservative frces, with what velocity does the rollercoaster reach the bottom of the drop?

B. After the largest drop the rollercoaster up another hill, this time 50m above the bottom. What is the velocity of the rollercoaster at the top of this second hill? Assuming no work is done by non-conservatice forces.

C. After the first two hills, the rollercoaster again reaches he bottom of it's path. Due to non-conservative forces (friction), the 800kg coaster is now moving at 45 m/s. At this point, how much work is done by friction?

Answer 1

From energy conservation,

KE + PE ( at height ) = KE + PE ( at bottom)

PE at the bottom most point is 0.

KE ( height ) = 0.5*m*v² = 2m

PE ( height ) = mgh = 1245.87m

Therefore, 2m + 1245.87m = 0.5*m*v² ===> v ( lowest point ) 49.9973 m/s

Now at the lowest point the velocity is 49.9973 m/s and the PE is 0.

By virtue of the KE at the lowest point, the ride will climb a 50 m hill ( gains PE) and also has some velocity at the top (KE)

Again using conservation of Energy

KE ( lowest point) = PE + KE ( 50 m height )

0.5*m*49.9973² = 490.5*m + 0.5*m*v² ====> v ( 50 m ) = 38.9708 m/s

In the last part, the roller coaster again comes back to the lowest point, so from the conservation of energy the coaster must have v = 49.9973 m/s

but it has the velocity of 45 m/s only.

Hence the loss in KE = 0.5*m* (49.9973² - 45² ) ===> 189892.0029 Joules

This loss in Energy is the work done by friction force, hence the answer is 189892.0029 Joules