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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.

Making connections: take-home investigation—converting potential to kinetic energy

One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see [link] ). Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot shows that the marble’s kinetic energy at the bottom is proportional to its potential energy at the release point.

A book is lying on the table and one end of a ruler rests on the edge of this book while the other end rests on the table, making it an incline. A marble is shown rolling down the ruler.
A marble rolls down a ruler, and its speed on the level surface is measured.

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

In [link] , we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy.

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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?

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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? (b) Compare this with the energy stored in a 9-megaton fusion bomb.

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"} {}

(b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb.

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(a) How much gravitational potential energy (relative to the ground on which it is built) is stored in the Great Pyramid of Cheops, given that its mass is about 7  ×  10 9  kg size 12{7 times "10" rSup { size 8{9} } " kg"} {} and its center of mass is 36.5 m above the surrounding ground? (b) How does this energy compare with the daily food intake of a person?

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Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch?

(a) 1.8 J

(b) 8.6 J

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In [link] , we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that Δ PE >> KE i size 12{D"PE>>KE" rSub { size 8{i} } } {} . Confirm this statement by taking the ratio of Δ PE size 12{D"PE"} {} to KE i size 12{"KE" rSub { size 8{i} } } {} . (Note that mass cancels.)

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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"} {}
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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.

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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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Practice Key Terms 1

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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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