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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.

Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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