Peoples Physics Book Version-2

2.5. Examples http://www.ck12.org

2.5 Examples

Example 1

QuestionA roller coaster begins at rest 120m above the ground at point A. Assume no energy is lost. The radius of
the loop is 40 ft

a) Find the speed of the roller coaster at point B, D, F, and H.

b) At point G the roller coaster’s speed is 22m/s. How high off the ground is point G?

FIGURE 2.1

Roller coaster for problem 8.

Solutiona) To solve for the speed at any point on the roller coaster, we use conservation of energy: the cart’s total
energy, equal to its initial energy (all potential) is split between kinetic and potential energy at all points during the
trip. Therefore, at any point

mgh+

1

2

mv^2 =mg× 120

Solving forv:

120 mg=

1

2

mv^2 +mgh⇒ 120 g−gh=

1

2

mv^2 ⇒g( 120 −h) =

1

2

mv^2 ⇒

√

2 g( 120 −h) =v

Therefore

B:

√

2 gh=

√

2 × 9 .8m/s^2 ×( 120 −60m) =34m/s

D:

√

2 gh=

√

2 × 9 .8m/s^2 ×( 120 −80m) =28m/s

F:

√

2 gh=

√

2 × 9 .8m/s^2 ×( 120 −0m) =48m/s

H:

√

2 gh=

√

2 × 9 .8m/s^2 ×( 120 −120m) =0m/s

b) As in part a), we start with the equation

mgh+

1

2

mv^2 =mg× 120

but this time we will solve for h.

120 mg=

1

2

mv^2 +mgh⇒ 120 g−

1

2

v^2 =gh⇒

120 g

g

−

v^2

2 g

=h⇒ 120 −

v^2

2 g

=h