**Getting the Coaster Started**

Something has to be done to get the coaster started. In our
previous example energy, power, has to added get the coaster up to
8.8 m/s. This is done by doing work on the coaster. A simplified
definition of work would be force times displacement when the force
and displacement go in the same direction. [This chapter will not go
into all the details of calculating work.] Suffice it to say that
when the force acting on the coaster and the displacement of the
coaster are in the same direction, work adds energy to the coaster.
When the force acting on the coaster and the displacement of the
coaster are in opposite directions, work removes energy from the
coaster.

Work = (Force)(Displacement)

W = Fd

Where "work" is measured in joules, J. "Force" is measured in
Newtons, N, and "displacement" is measured in meters, m.

Example 1

What is the velocity of the train after being catapulted into
motion?

Solution:

ET_{(BEGINNING) + Work }= ET_{(TOP OF 1}^{st}_{ HILL)}

KE + PE +W = KE +PE

(^{1}/2)mv^{2} + mgh + Fd =
(^{1}/2)mv^{2} + mgh

(^{1}/2)3000(0)^{2} +
3000(9.8)(3000) + 8800(12.5)= (^{1}/2)3000v^{2} + 3000(9.8)(0) Substitute
the numbers at each

location

110,000 = (^{1}/2)(3000)v^{2
}73.333 = v^{2}

v = 8.6 ^{m}/s
... at the end of the catapult.

As an aside you can calculate the acceleration of the rider from
kinematics equations.

(For the curious the acceleration is 7.0 ^{m}/s^{2})^{
}Example 2

What is the velocity of the train after being catapulted into
motion?

^{
`
}Solution:

ET_{(BEGINNING) + Work }= ET_{(TOP OF 1}^{st}_{ HILL)}

KE + PE +W = KE +PE

(^{1}/2)mv^{2} + mgh + Fd =
(^{1}/2)mv^{2} + mgh

(^{1}/2)3000(5.2)^{2} +
3000(9.8)(0) + 19,715.464(145.5) = (^{1}/2)3000v^{2} + 3000(9.8)(95)

40560 + 2868600.012 = (^{1}/2)(3000)v^{2 }+
2793000^{
}116160.012 = (^{1}/2)(3000)v^{2
}77.44 =^{ }v^{2}

v = 8.8 ^{m}/s
... at the end of the catapult.

Example 3

A roller coaster train of mass 3.0 X 10^{3} kg rolls over a 11.5 m high hill at 8.34
^{m}/s before
rolling down into the station. Once in the station, brakes are
applied to the train to slow it down to 1.00 ^{m}/s in 5.44 m.

(a) What braking force slowed the train down?

(b) How much time did it take to slow the train down?

(c) What was the acceleration of the train in g's?

Solution:

(a)

ET_{(HILL) }= ET_{(@ 1 m/s)} _{+ Work}

KE + PE = KE +PE +W

(^{1}/2)mv^{2} + mgh =
(^{1}/2)mv^{2} + mgh +
Fd

(^{1}/2)3000(8.34)^{2} +
3000(9.8)(11.5) = (^{1}/2)3000(1)^{2} +
3000(9.8)(0) + F(5.44) Substitute the numbers

at each location

442433.400 = 1500 + 5.44F^{
}__F = 81053.9__ N ...force to slow down the train

(b)

Calculate the velocity as the train enters the station. Use this
velocity to calculate the time.

ET_{(HILL) }= ET_{(@ STATION ENTRANCE)} No work is done because
no force acts between the two locations

KE + PE = KE +PE

(^{1}/2)mv^{2} + mgh =
(^{1}/2)mv^{2} + mgh

(^{1}/2)3000(8.34)^{2} +
3000(9.8)(11.5) = (^{1}/2)3000v^{2} +
3000(9.8)(0) Substitute the numbers

at each location

442433.400 = 1500v^{2}

v = 17.174

__v = 17.2__ ^{m}/s ...as the train enters the station.^{
}

The time is calculated from

v_{o} = 17.174 m/s

v_{f} = 1.00 m/s

x = 5.44 m

t = ?

__t = 0.599 sec__

(c)

Calculate the acceleration in ^{m}/s^{2}. Then convert it into g's.

F= ma

81053.934 N = (3000 kg)a

a = 37.018 ^{m}/s^{2
}a = 37.018 ^{m}/s^{2}/ 9.80
^{m}/s^{2} = __3.78
g__'__s__ ... Yeow! That's a big jerk on the passengers

into the restraining harness.

If you use or find this page useful or have any
comments, please contact the author so maybe he'll do more.
Author: Tony Wayne

"ROLLER COASTER PHYSICS" TABLE
OF CONTENTS ... PHYSICS PAVILION TABLE OF
CONTENTS

<--PREVIOUS
SECTION ... NEXT SECTION
-->