(a) What is the final speed of the roller coaster shown in Figure 4 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s?

*Strategy*

The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The *loss* of gravitational potential energy from moving *downward* through a distance hh size 12{h} {} equals the *gain* in kinetic energy. This can be written in equation form as −ΔPEg=ΔKE−ΔPEg=ΔKE size 12{ - Δ"PE" rSub { size 8{g} } =Δ"KE"} {}. Using the equations for PEgPEg size 12{"PE" rSub { size 8{g} } } {} and KEKE size 12{"KE"} {}, we can solve for the final speed vv size 12{v} {}, which is the desired quantity.

*Solution for (a)*

Here the initial kinetic energy is zero, so that ΔKE=12mv2ΔKE=12mv2. The equation for change in potential energy states that ΔPEg=mghΔPEg=mgh. Since hh is negative in this case, we will rewrite this as ΔPEg=−mg∣h∣ΔPEg=−mg∣h∣ to show the minus sign clearly. Thus,

−
Δ
PE
g
=
Δ
KE
−
Δ
PE
g
=
Δ
KE
size 12{ - Δ"PE" rSub { size 8{g} } =Δ"KE"} {}

(8)becomes

mg
∣
h
∣
=
1
2
mv
2
.
mg
∣
h
∣
=
1
2
mv
2
.
size 12{ ital "mg" lline h rline = { {1} over {2} } ital "mv" rSup { size 8{2} } "." } {}

(9)Solving for vv size 12{v} {}, we find that mass cancels and that

v=2g∣h∣.v=2g∣h∣. size 12{v= sqrt {2g lline h rline } } {}

(10)Substituting known values,

v
=
2
9
.
80 m
/s
2
20.0 m
=
19
.8 m/s.
v
=
2
9
.
80 m
/s
2
20.0 m
=
19
.8 m/s.
alignl { stack {
size 12{v= sqrt {2 left (9 "." "80"" m/s" rSup { size 8{2} } right ) left ("20" "." 0" m" right )} } {} #
" "=" 19" "." "8 m/s" "." {}
} } {}

(11)*Solution for (b)*

Again −ΔPEg=ΔKE−ΔPEg=ΔKE size 12{ - Δ"PE" rSub { size 8{g} } =Δ"KE"} {}. In this case there is initial kinetic energy, so ΔKE=12
m
v
2
−12m
v
0
2
ΔKE=12
m
v
2
−12m
v
0
2
size 12{Δ"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } } {}. Thus,

mg
∣
h
∣
=
1
2
mv
2
−
1
2
m
v
0
2
.
mg
∣
h
∣
=
1
2
mv
2
−
1
2
m
v
0
2
.
size 12{ ital "mg" lline h rline = { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } "." } {}

(12)Rearranging gives

1
2
mv
2
=
mg
∣
h
∣
+
1
2
m
v
0
2
.
1
2
mv
2
=
mg
∣
h
∣
+
1
2
m
v
0
2
.
size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } = ital "mg" lline h rline + { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } "." } {}

(13)This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and

v=2g∣h∣+
v
0
2
.v=2g∣h∣+
v
0
2
. size 12{v= sqrt {2g lline h rline +v rSub { size 8{0} rSup { size 8{2} } } } } {}

(14)This equation is very similar to the kinematics equation v=
v
0
2
+2adv=
v
0
2
+2ad size 12{v= sqrt {v rSub { size 8{0} } rSup { size 8{2} } +2 ital "ad"} } {}, but it is more general—the kinematics equation is valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives

v
=
2
(
9
.
80
m/s
2
)
(
20
.0 m
)
+
(
5
.00 m/s
)
2
=
20.4 m/s.
v
=
2
(
9
.
80
m/s
2
)
(
20
.0 m
)
+
(
5
.00 m/s
)
2
=
20.4 m/s.
alignl { stack {
size 12{v= sqrt {2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) \( "20" "." 0" m" \) + \( 5 "." "00"" m/s" \) rSup { size 8{2} } } } {} #
" "=" 20" "." "4 m/s" "." {}
} } {}

(15)*Discussion and Implications*

First, note that mass cancels. This is quite consistent with observations made in Falling Objects that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, note that speed can be found at *any* height along the way by simply using the appropriate value of hh size 12{h} {} at the point of interest.

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