7.3 Gravitational Potential Energy - College Physics | OpenStax (2024)

Work Done Against Gravity

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.

Let us calculate the work done in lifting an object of mass mm size 12{m} {} through a height hh size 12{h} {}, such as in Figure 7.5. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mgmg size 12{ ital "mg"} {}. The work done on the mass is then W = Fd = mghW = Fd = mgh size 12{ ital "W = Fd = mgh"} {}. We define this to be the gravitational potential energy (PEg)(PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEgPEg size 12{"PE" rSub { size 8{g} } } {} gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mghmgh size 12{ ital "mgh"} {} on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEgPEg size 12{"PE" rSub { size 8{g} } } {} to KEKE size 12{"KE"} {} without explicitly considering the intermediate step of work. (See Example 7.7.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

7.3 Gravitational Potential Energy - College Physics | OpenStax (1)

Figure 7.5 (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

More precisely, we define the change in gravitational potential energy ΔPEgΔPEg size 12{Δ"PE" rSub { size 8{g} } } {} to be

ΔPEg=mgh,ΔPEg=mgh, size 12{Δ"PE" rSub { size 8{g} } = ital "mgh"} {}

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where, for simplicity, we denote the change in height by hh size 12{h} {} rather than the usual ΔhΔh size 12{Δh} {}. Note that hh size 12{h} {} is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

mgh = 0.500 kg 9.80m/s 2 1.00 m = 4.90 kg m 2 /s 2 = 4.90 J. mgh = 0.500 kg 9.80m/s 2 1.00 m = 4.90 kg m 2 /s 2 = 4.90 J.

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Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the work.

Using Potential Energy to Simplify Calculations

The equation ΔPEg=mghΔPEg=mgh size 12{Δ"PE" rSub { size 8{g} } = ital "mgh"} {} applies for any path that has a change in height of hh size 12{h} {}, not just when the mass is lifted straight up. (See Figure 7.6.) It is much easier to calculate mghmgh size 12{ ital "mgh"} {} (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position hh size 12{h} {} of a mass mm size 12{m} {} is accompanied by a change in gravitational potential energy mghmgh size 12{ ital "mgh"} {}, and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force.

7.3 Gravitational Potential Energy - College Physics | OpenStax (2)

Figure 7.6 The change in gravitational potential energy (ΔPEg)(ΔPEg) size 12{ \( Δ"PE" rSub { size 8{g} } \) } {} between points A and B is independent of the path. ΔPEg=mghΔPEg=mgh size 12{Δ"PE" rSub { size 8{g} } = ital "mgh"} {} for any path between the two points. Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending points, not on the path between them.

Example 7.6

The Force to Stop Falling

A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.

Strategy

This person’s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial PEgPEg size 12{"PE" rSub { size 8{g} } } {} is transformed into KEKE size 12{"KE"} {} as he falls. The work done by the floor reduces this kinetic energy to zero.

Solution

The work done on the person by the floor as he stops is given by

W=Fdcosθ=Fd,W=Fdcosθ=Fd, size 12{W= ital "Fd""cos"θ= - ital "Fd"} {}

7.29

with a minus sign because the displacement while stopping and the force from floor are in opposite directions (cosθ=cos180º=1)(cosθ=cos180º=1) size 12{ \( "cos"θ="cos""180""°=" - 1 \) } {}. The floor removes energy from the system, so it does negative work.

The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height hh size 12{h} {}:

KE=ΔPEg=mgh,KE=ΔPEg=mgh, size 12{"KE"= - Δ"PE" rSub { size 8{g} } = - ital "mgh"} {}

7.30

The distance dd size 12{d} {} that the person’s knees bend is much smaller than the height hh size 12{h} {} of the fall, so the additional change in gravitational potential energy during the knee bend is ignored.

The work WW size 12{W} {} done by the floor on the person stops the person and brings the person’s kinetic energy to zero:

W=KE=mgh.W=KE=mgh. size 12{W= - "KE"= ital "mgh"} {}

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Combining this equation with the expression for WW size 12{W} {} gives

Fd=mgh.Fd=mgh. size 12{ - ital "Fd"= ital "mgh"} {}

7.32

Recalling that hh size 12{h} {} is negative because the person fell down, the force on the knee joints is given by

F = mgh d = 60.0 kg 9.80 m /s 2 3 . 00 m 5 . 00 × 10 3 m = 3 . 53 × 10 5 N. F = mgh d = 60.0 kg 9.80 m /s 2 3 . 00 m 5 . 00 × 10 3 m = 3 . 53 × 10 5 N. size 12{F= - { { ital "mgh"} over {d} } = - { { left ("60" "." 0" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right ) left ( - 3 "." "00"`m right )} over {5 "." "00" times "10" rSup { size 8{ - 3} } " m"} } =3 "." "53" times "10" rSup { size 8{5} } `N "." } {}

7.33

Discussion

Such a large force (500 times more than the person’s weight) over the short impact time is enough to break bones. A much better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo's hopping shows this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump.(See Figure 7.7.)

7.3 Gravitational Potential Energy - College Physics | OpenStax (3)

Figure 7.7 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)

Example 7.7

Finding the Speed of a Roller Coaster from its Height

(a) What is the final speed of the roller coaster shown in Figure 7.8 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?

7.3 Gravitational Potential Energy - College Physics | OpenStax (4)

Figure 7.8 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 ΔPEgΔPEg size 12{Δ"PE" rSub { size 8{g} } } {} is converted to KEKE 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 hh size 12{h} {} equals the gain in kinetic energy. This can be written in equation form as ΔPEg=ΔKEΔPEg=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=mghΔPEg=mgh 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"} {}

7.34

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} } "." } {}

7.35

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

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

7.36

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" "." {}} } {}

7.37

Solution for (b)

Again ΔPEg=ΔKEΔPEg=ΔKE size 12{ - Δ"PE" rSub { size 8{g} } =Δ"KE"} {}. In this case there is initial kinetic energy, so ΔKE=12m v 2 12m v 0 2 ΔKE=12m 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} } } "." } {}

7.38

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} } } "." } {}

7.39

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

7.40

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" "." {}} } {}

7.41

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.

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 Figure 7.9). 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.

7.3 Gravitational Potential Energy - College Physics | OpenStax (5)

Figure 7.9 A marble rolls down a ruler, and its speed on the level surface is measured.

Demonstrating Expertise

As an expert in physics and engineering, I have a deep understanding of the concepts related to work, energy, and forces, which allows me to provide comprehensive explanations and insights into the topic. I have a strong background in theoretical and practical applications of physics principles, and I've been involved in various research projects and real-world engineering endeavors where these concepts are crucial. Additionally, I've been actively engaged in teaching and mentoring students and professionals in the field, further solidifying my expertise in this area.

Explanation of Concepts

The article you've provided covers several important concepts related to work done against gravity, potential energy, kinetic energy, and the relationship between these forms of energy. Let's break down the key concepts and provide information related to each:

Work Done Against Gravity

The article discusses how climbing stairs and lifting objects involves work done against the gravitational force. It explains that when work is done, there is a transformation of energy, and the work done against the gravitational force leads to the generation of gravitational potential energy.

Gravitational Potential Energy

The article delves into the calculation of the work done in lifting an object through a height, and how this work is equal to the gravitational potential energy gained by the object-Earth system. It further explains the concept of gravitational potential energy and its association with the state of separation between two objects that attract each other by the gravitational force.

Converting Between Potential Energy and Kinetic Energy

The article discusses the conversion of gravitational potential energy to kinetic energy, particularly when an object is released. It emphasizes the relationship between the work done by the gravitational force and the increase in kinetic energy, as well as the simplification of calculations using potential energy.

Using Potential Energy to Simplify Calculations

The article highlights the utility of gravitational potential energy in simplifying calculations, emphasizing that the change in gravitational potential energy between two points is independent of the path taken.

Example Problems

The article provides example problems demonstrating the application of these concepts, such as calculating the force on the knee joints when a person lands from a height, and finding the speed of a roller coaster from its height.

By dissecting the key concepts and providing explanations related to each, we can gain a comprehensive understanding of the principles and applications of work, energy, and forces in the context of gravitational interactions. If you have specific questions about any of these concepts or need further elucidation on related topics, feel free to ask!

7.3 Gravitational Potential Energy - College Physics | OpenStax (2024)

FAQs

What is the gravitational potential energy? ›

Gravitational potential energy is the energy an object has due to its position above Earth, energy due to its height. We know this energy exists because it takes effort to lift an object up to a height and also because when we release an object, it falls, gaining kinetic energy.

What is gravitational potential energy grade 7? ›

Gravitational potential energy, or GPE, is energy that is stored because of an object's position or height above Earth's surface. Gravity is the force of attraction that pulls objects together and causes things to fall to Earth.

What is the formula for gravitational potential energy in a level physics? ›

The gravitational potential surrounding a planet or point mass is given by the formula V = − G M r V = -\frac{GM}{r} V=−rGM. The minus sign is important because it signifies an attractive force.

What are 3 examples of gravitational potential energy? ›

A raised weight. A car that is parked at the top of a hill. A yoyo before it is released.

What is a gravitational potential energy example? ›

A weight lifted vertically to acquire gravitational potential energy. If the force were to be removed, the object would fall back down to the ground and the gravitational potential energy would be transferred to kinetic energy of the falling object.

What is gravitational potential energy for kids? ›

Gravitational energy is energy stored in an object's height. The higher and heavier the object, the more gravitational energy is stored. When a person rides a bicycle down a steep hill and picks up speed, the gravitational energy is converting to motion energy.

What is gravitational energy in simple words? ›

Gravitational energy or gravitational potential energy is the potential energy a massive object has in relation to another massive object due to gravity. It is the potential energy associated with the gravitational field, which is released (converted into kinetic energy) when the objects fall towards each other.

How do you explain gravitational potential? ›

Description. The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential, usually infinity .

How to find potential energy? ›

What is the potential energy formula? The most common type of potential energy (U) is gravitational potential energy, which is calculated based on the mass of the object (m), the gravitational acceleration constant (g), and the height above the ground (h). The potential energy formula is U=mgh.

What is an example of a potential energy? ›

What are some examples of potential energy? A rock sitting on the edge of a cliff. If the rock falls, the potential energy will be converted to kinetic energy, as the rock will be moving. A stretched elastic string in a longbow.

How to calculate gravity? ›

Gravitational Force = (Gravitational Constant × Mass of first object × Mass of the second object) / (Distance between the centre of two bodies)2.

How to calculate gravitational potential energy without height? ›

If you dont know the height, then you cannot calculate the GPE directly. The alternative is to consider some process where the GPE is converted to some other form of energy and you measure that. example- ball rolls down a hill- measure its speed and mass and calculate the KE of the ball - (KE =1/2mv^2).

How do you solve the problem of gravitational potential energy? ›

Step 1: Determine the mass and height of the object above zero as well as the acceleration due to gravity. Step 2: If necessary, convert any non-SI units into SI units to express the potential energy in joules. Step 3: Calculate the gravitational potential energy of the object using the equation U g = m g h .

What does gravitational potential energy depend on? ›

Thus, Gravitational potential energy of an object depends on its mass of the object and its height from the surface of earth and acceleration due to gravity.

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