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A hopping kangaroo is shown landing on the ground in one photograph and in the air just after taking another jump in the second photograph.
The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground on the hind legs over a longer distance, the impact on the bones is reduced. (credit: Chris Samuel, Flickr)

Finding the speed of a roller coaster from its height

(a) What is the final speed of the roller coaster shown in [link] 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?

A roller coaster track is shown with a car about to go downhill. The initial height of the roller coaster car on the track is twenty-five meters from the lowest part of the track and its speed v sub zero is equal to zero. The roller coaster’s height from the level part of the track is twenty meters. The finish point of the car is on the level part of the track and the speed at that point is unknown.
The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system’s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all Δ PE g size 12{Δ"PE" rSub { size 8{g} } } {} is converted to KE size 12{"KE"} {} .

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 h size 12{h} {} equals the gain in kinetic energy. This can be written in equation form as Δ PE g = Δ KE size 12{ - Δ"PE" rSub { size 8{g} } =Δ"KE"} {} . Using the equations for PE g size 12{"PE" rSub { size 8{g} } } {} and KE size 12{"KE"} {} , we can solve for the final speed v size 12{v} {} , which is the desired quantity.

Solution for (a)

Here the initial kinetic energy is zero, so that ΔKE = 1 2 mv 2 . The equation for change in potential energy states that ΔPE g = mgh . Since h is negative in this case, we will rewrite this as ΔPE g = mg h to show the minus sign clearly. Thus,

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

becomes

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

Solving for v size 12{v} {} , we find that mass cancels and that

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

Substituting known values,

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" "." {} } } {}

Solution for (b)

Again ΔPE g = ΔKE size 12{ - Δ"PE" rSub { size 8{g} } =Δ"KE"} {} . In this case there is initial kinetic energy, so ΔKE = 1 2 m v 2 1 2 m 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 . 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} } } "." } {}

Rearranging gives

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} } } "." } {}

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

This equation is very similar to the kinematics equation v = v 0 2 + 2 ad 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. 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" "." {} } } {}

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 h size 12{h} {} at the point of interest.

We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy.

Section summary

  • Work done against gravity in lifting an object becomes potential energy of the object-Earth system.
  • The change in gravitational potential energy, Δ PE g size 12{Δ"PE" rSub { size 8{g} } } {} , is ΔPE g = mgh size 12{Δ"PE" rSub { size 8{g} } = ital "mgh"} {} , with h size 12{h} {} being the increase in height and g size 12{g} {} the acceleration due to gravity.
  • The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, ΔPE g size 12{Δ"PE" rSub { size 8{g} } } {} , have physical significance.
  • As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that ΔKE = − ΔPE g size 12{D"KE""=-"D"PE" rSub { size 8{g} } } {} .

Conceptual questions

Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the mass of the book?

Problems&Exercises

A hydroelectric power facility (see [link] ) converts the gravitational potential energy of water behind a dam to electric energy. (a) What is the gravitational potential energy relative to the generators of a lake of volume 50 . 0 k m 3 size 12{"50" "." 0`"km" rSup { size 8{3} } } {} ( mass = 5 . 00 × 10 13 kg ) size 12{"mass"=5 "." "00" times "10" rSup { size 8{"13"} } `"kg " \) } {} , given that the lake has an average height of 40.0 m above the generators?

A dam with water flowing down its gates.
Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons)

(a) 1 . 96 × 10 16 J size 12{1 "." "96" times "10" rSup { size 8{"16"} } " J"} {}

A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in [link] . Show that the final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude.

A toy car is moving up a curved track.
A toy car moves up a sloped track. (credit: Leszek Leszczynski, Flickr)
v f = 2 gh + v 0 2 = 2 ( 9.80 m /s 2 ) ( 0 .180 m ) + ( 2 .00 m/s ) 2 = 0 .687 m/s size 12{v rSub { size 8{f} } = sqrt {2 ital "gh"+v rSub { size 8{0} rSup { size 8{2} } } } = sqrt {2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) \( - 0 "." "180"" m" \) + \( 2 "." "00 m/s" \) rSup { size 8{2} } } =0 "." "687"" m/s"} {}

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a 30º size 12{"30"°} {} slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events.

Questions & Answers

anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Unit 5 - work and energy. OpenStax CNX. Jan 02, 2016 Download for free at https://legacy.cnx.org/content/col11946/1.1
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