<< Chapter < Page Chapter >> Page >
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.

Questions & Answers

material that allows electric current to pass through
Deng Reply
material which don't allow electric current is called
Deng
insulators
Covenant
how to study physic and understand
Ewa Reply
what is conservative force with examples
Moses
what is work
Fredrick Reply
the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement
AI-Robot
why is it from light to gravity
Esther Reply
difference between model and theory
Esther
Is the ship moving at a constant velocity?
Kamogelo Reply
The full note of modern physics
aluet Reply
introduction to applications of nuclear physics
aluet Reply
the explanation is not in full details
Moses Reply
I need more explanation or all about kinematics
Moses
yes
zephaniah
I need more explanation or all about nuclear physics
aluet
Show that the equal masses particles emarge from collision at right angle by making explicit used of fact that momentum is a vector quantity
Muhammad Reply
yh
Isaac
A wave is described by the function D(x,t)=(1.6cm) sin[(1.2cm^-1(x+6.8cm/st] what are:a.Amplitude b. wavelength c. wave number d. frequency e. period f. velocity of speed.
Majok Reply
what is frontier of physics
Somto Reply
A body is projected upward at an angle 45° 18minutes with the horizontal with an initial speed of 40km per second. In hoe many seconds will the body reach the ground then how far from the point of projection will it strike. At what angle will the horizontal will strike
Gufraan Reply
Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen? Such differences in arrival times are used as an analytical tool in gas chromatography.
Ezekiel Reply
please explain
Samuel
what's the definition of physics
Mobolaji Reply
what is physics
Nangun Reply
the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental mechanisms underlie every phenomenon
AI-Robot
Practice Key Terms 1

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask