Lesson Explainer: Potential Energy Mathematics

In this explainer, we will learn how to find potential energy and the change in it and use it to solve different problems.

Imagine an object that is held at rest at a height above the ground and is then allowed to fall. As the object falls, it will accelerateβ€”its speed will increase, and so will its kinetic energy.

We know, however, that energy is conserved. Energy cannot be created or destroyed, it can only be transferred. This means that in order for the kinetic energy of the object to increase, another energy must decrease by the same amount. For the falling object, this is its gravitational potential energy.

All objects in the universe that have mass attract each other. The force of gravity acts between objects with mass to move them closer together, increasing their speed and their kinetic energy.

This means that when the objects are separated, we can say that they have potential energyβ€”in this case specifically, gravitational potential energy. Gravitational potential energy exists for any system of massive objects that are separated.

Let us consider a body falling on a smooth inclined plane from point 𝐴 to point 𝐡. Friction forces are neglected, and so, the object is acted upon only by the gravitational force, its weight, given by π‘šβƒ‘π‘”.

According to the work–energy principle, the change in kinetic energy when the body falls from 𝐴 to 𝐡 is given by the work, π‘Š, of the body’s weight between 𝐴 and 𝐡: Δ𝐸=π‘Š=ο„Έπ‘šβƒ‘π‘”β‹…βƒ‘π‘ .Kd

For any infinitesimal displacement d⃑𝑠, π‘šβƒ‘π‘”β‹…βƒ‘π‘ d can be interpreted as the scalar projection of d⃑𝑠 on ⃑𝑔 multiplied by π‘šπ‘”, with 𝑔 being the magnitude of ⃑𝑔. The integral ο„Έπ‘šβƒ‘π‘”β‹…βƒ‘π‘ οŒ‘οŒ d is thus the sum of these scalar projections, that is, the projection of 𝐴𝐡 on ⃑𝑔, multiplied by π‘šπ‘” (i.e., π‘šβƒ‘π‘”β‹…οƒ«π΄π΅). Since ⃑𝑔 is a vertical vector pointing down, the scalar projection of 𝐴𝐡 on ⃑𝑔 is the opposite of the change in the 𝑧-coordinate (also called the height, or altitude) from 𝐴 to 𝐡 (since it is positiveβ€”in the same direction of ⃑𝑔—when the altitude decreases). Hence, we have π‘Š=ο„Έπ‘šβƒ‘π‘”β‹…βƒ‘π‘ =βˆ’π‘šπ‘”(π‘§βˆ’π‘§)=βˆ’π‘šπ‘”Ξ”β„Ž,d where π‘§οŒ  and π‘§οŒ‘ are the 𝑧-coordinates of 𝐴 and 𝐡, respectively, and Ξ”β„Ž is the change in altitude from 𝐴 to 𝐡.

The gravitational potential energy decreases by the same amount that the kinetic energy increases by when a bodyβ€”subjected only to its weightβ€”falls, since the body’s total energy is conserved. Therefore, with Δ𝐸G the change in gravitational potential energy, we have Δ𝐸+Δ𝐸=0;KG that is, Δ𝐸=βˆ’Ξ”πΈ=βˆ’π‘Š.GK

This means that, inversely, when the kinetic energy decreases, the potential energy increases.

The two situations are illustrated with a train on a roller coaster: when it goes down a drop, it loses potential energy as it loses altitude and gains kinetic energy; when it then goes up the next drop, its kinetic energy decreasesas it gains height (and thus potential energy).

The gravitational potential energy of an object is dependent on the mass and the height of the object and is defined as follows.

Definition: Gravitational Potential Energy

The gravitational potential energy, 𝐸G, of an object with a mass π‘š that is a height β„Ž above a reference point is given by 𝐸=π‘šπ‘”β„Ž,G where 𝑔 is the local acceleration due to gravity with a value of approximately 9.8 m/s2.

Note that β„Ž can be the height above any reference point we choose. We could choose the top of a building or the top of a cliff, in which case the gravitational potential energy that we calculate would be the potential energy of the object relative to the top of the building or the top of the cliff. Most of the time, however, the reference point we choose is the ground, in which case objects that are on the ground have a gravitational potential energy of zero.

Let us have a look at our first example where we need to find the height of a body given its mass and potential energy.

Example 1: Determining the Height above the Ground of a Body Using Gravitational Potential Energy

A body of mass 4 kg had a gravitational potential energy of 2β€Žβ€‰β€Ž136.4 joules relative to the ground. Determine its height. Consider the acceleration due to gravity to be 9.8 m/s2.

Answer

To solve this question, we can use the formula for the gravitational potential energy of an object: 𝐸=π‘šπ‘”β„Ž.G

First, let’s rearrange the formula to make β„Ž the subject: πΈπ‘šπ‘”=β„Žβ„Ž=πΈπ‘šπ‘”.GG

Now, let’s substitute in the values: β„Ž=2136.44Γ—9.8β„Ž=54.5.

The object is at a height of 54.5 m above the ground.

Let us have a look at an example where we need to find the increase in the potential energy of a body that is lifted up.

Example 2: Finding the Gravitational Potential Energy of a Body at a Given Height above the Ground

A crane lifts a body of mass 132 kg to a height of 20 m. Find the increase in the body’s gravitational potential energy. Consider the acceleration due to gravity 𝑔=9.8/ms.

Answer

In the scenario in this question, a crane lifts an object up to a height of 20 m. The question does not say where the object startsβ€”it might start on the ground or it might have already been on part of a structure higher upβ€”but it does not matter because we can take the reference point to be wherever the object started and we only need to consider how much higher the object is after it has been moved compared to before it was moved, which is 20 m.

We can use the formula for the gravitational potential energy of an object:𝐸=π‘šπ‘”β„Ž.G

Let’s substitute in the values: 𝐸=132Γ—9.8/Γ—20𝐸=25872β‹…/.GGkgmsmkgms

Units of kilogram-square metres per second squared (kgβ‹…m2/s2) are equivalent to units of joules (J) so the increase in the gravitational potential energy of the object is 25β€Žβ€‰β€Ž872 J.

Our next example deals with a body moving up an inclined plane.

Example 3: Finding the Increase in the Gravitational Potential Energy of a Body Moving up an Inclined Plane

A body of mass 8 kg moved 238 cm up the line of greatest slope of a smooth plane inclined at 30∘ to the horizontal. Calculate the increase in its gravitational potential energy. Take 𝑔=9.8/ms.

Answer

This problem is easier to solve if we first draw a diagram of the scenario.

Here we’re measuring the distance moved up the slope and the increase in height of the lower-left corner of the object, but, since the object does not change shape as it moves, we could have chosen any point.

The object moves 238 cm up the slope, but we are only interested in its increase in height since the gravitational potential energy of an object is dependent on its height, not the distance moved.

We can use trigonometry to find the increase in height, β„Ž, of the object: β„Ž=238(30)sin, which is equal to 119 cm or 1.19 m. We will use metres in our calculation so that we get a value for the energy in joules.

Now we can use the formula for the gravitational potential energy of an object: 𝐸=π‘šπ‘”β„Ž.G

Let’s substitute in the values: 𝐸=8Γ—9.8Γ—1.19𝐸=93.296.GG

So, the increase in the object’s gravitational potential energy is 93.296 joules.

Let us now look at an example of loss in gravitational energy.

Example 4: Finding the Change in the Gravitational Potential Energy of a Helicopter Descending Vertically

A helicopter of mass 2β€Žβ€‰β€Ž630 kg descended vertically from a height of 250 m to a height of 150 m. Find its loss in gravitational potential energy. Consider the acceleration due to gravity to be 𝑔=9.8/ms and give your answer in scientific notation.

Answer

In order to find the amount by which the gravitational potential energy of the helicopter decreases, we must find the difference in the gravitational potential energy of the helicopter at the two heights. This can be expressed algebraically as Δ𝐸=π‘šπ‘”β„Žβˆ’π‘šπ‘”β„ŽΞ”πΈ=π‘šπ‘”(β„Žβˆ’β„Ž),GG where β„ŽοŠ¨ is the upper height and β„ŽοŠ§ is the lower height. Let’s substitute in the values: Δ𝐸=2630Γ—9.8Γ—(250βˆ’150)Δ𝐸=2630Γ—9.8Γ—100Δ𝐸=2577400.GGG

This answer is in joules and can be expressed in scientific notation as 2.5774Γ—10J or 2.6Γ—10J to one decimal place.

So far, we have looked only at the example of gravitational potential energy caused by the gravitational force. There are other forces that lead to potential energy, such as the elastic force in a spring (leading to the elastic potential energy) or the electric force on an electric charge placed in an electric field (leading to the electric potential energy). All these forces are said to be conservative because their work does not depend on the path taken, and so, if a body travels in a closed loop, the total work is zero.

As we have seen above for the gravitational force, it happens for instance when the force is constant because its dot product with an infinitesimal displacement d⃑𝑠 is proportional to the scalar projection of d⃑𝑠 on a constant vector (the force vector). The integral to find the work of the force when the body goes from a point to another will therefore not depend on the path taken but only on the initial and end positions.

The concept of potential energy is then extended to any conservative force and the change in potential energy due to the action of a conservative force is defined as follows.

Definition: Change in Potential Energy due to a Conservative Force

The change in the potential energy, Δ𝐸P, of a body due to the action of a conservative force, ⃑𝐹, moving the body from point 𝐴 to point 𝐡 is the negative of the work done by the force between these two points: Δ𝐸=βˆ’π‘Š=βˆ’βƒ‘πΉβ‹…οƒ«π΄π΅.P

Let us look at an example where we need to find the change in the potential energy of a body moving under the action of a constant force.

Example 5: Finding the Change in Potential Energy due to a Constant Force

A body is moving under the action of a constant force ⃑𝐹=ο€Ί5⃑𝑖+3⃑𝑗N, where ⃑𝑖 and ⃑𝑗 are two perpendicular unit vectors. At time 𝑑 seconds, where 𝑑β‰₯0, the body’s position vector relative to a fixed point is given by βƒ‘π‘Ÿ=𝑑+4⃑𝑖+ο€Ή4𝑑+8ο…βƒ‘π‘—ο†οŠ¨οŠ¨m. Determine the change in the body’s potential energy in the first 9 seconds.

Answer

Because this is a constant force, it is a conservative force. For a conservative force, the change in an object’s potential energy, π‘ˆ, is given by Ξ”π‘ˆ=βˆ’π‘Š, where π‘Š is the work done. We can find the work done using π‘Š=⃑𝐹⋅⃑𝑑, where ⃑𝐹 is the force on the object and ⃑𝑑 is its displacement.

Since the force is a conservative force, the exact path taken by the object does not matter when calculating the work done; all we need to consider is the difference between the initial and final points in the object’s path. So, the displacement is just the final position vector of the object, βƒ‘π‘Ÿ(𝑑), minus the initial position vector, βƒ‘π‘Ÿ(𝑑): ⃑𝑑=βƒ‘π‘Ÿ(𝑑)βˆ’βƒ‘π‘Ÿ(𝑑)⃑𝑑=βƒ‘π‘Ÿ(9)βˆ’βƒ‘π‘Ÿ(0)⃑𝑑=ο€Ίο€Ή9+4⃑𝑖+ο€Ή4Γ—9+8ο…βƒ‘π‘—ο†βˆ’ο€Ίο€Ή0+4⃑𝑖+ο€Ή4Γ—0+8⃑𝑗⃑𝑑=ο€Ί85⃑𝑖+332βƒ‘π‘—ο†βˆ’ο€Ί4⃑𝑖+8⃑𝑗⃑𝑑=81⃑𝑖+324⃑𝑗.

We can now use the above formula to calculate the work done by the force: π‘Š=ο€Ί5⃑𝑖+3⃑𝑗⋅81⃑𝑖+324βƒ‘π‘—ο†π‘Š=1377.

Therefore, the change in potential energy is Ξ”π‘ˆ=βˆ’π‘ŠΞ”π‘ˆ=βˆ’1377.J

Let us look at our final example where we consider a conservative force.

Example 6: Finding the Unknown Component of a Force Acting on a Body given the Change in Its Potential Energy

A body is moving in a straight line from point 𝐴(βˆ’6,0) to point 𝐡(βˆ’5,4) under the action of the force ⃑𝐹=ο€Ίπ‘šβƒ‘π‘–+2⃑𝑗N. Given that the change in the body’s potential energy is 2 joules and that the displacement is in metres, determine the value of the constant π‘š.

Answer

In this question, we know that a force acts on an object as it moves along a path. The force does work on the object, and the value of the work done on the object is equal to the negative change in its potential energy.

We can use the formula for the work done on an object in order to come up with an expression that includes the unknown constant π‘š: π‘Š=⃑𝐹⋅⃑𝑑.

First, let’s find ⃑𝑑, the displacement that the object undergoes. We will assume that the coordinates given to us are in units of meters. We can represent points 𝐴 and 𝐡 as vectors βƒ‘π‘Ž and ⃑𝑏 with components βƒ‘π‘Ž=βˆ’6⃑𝑖 and ⃑𝑏=βˆ’5⃑𝑖+4⃑𝑗. The displacement, ⃑𝑑, is then given by ⃑𝑑=βƒ‘π‘βˆ’βƒ‘π‘Žβƒ‘π‘‘=ο€Ίβˆ’5⃑𝑖+4βƒ‘π‘—ο†βˆ’ο€Ίβˆ’6⃑𝑖⃑𝑑=⃑𝑖+4⃑𝑗.

Now, let’s put the expressions for the force and the displacement, as well as the value of the work done, into the formula for the work: Ξ”π‘ˆ=βˆ’π‘Šβˆ’2=ο€Ίπ‘šβƒ‘π‘–+2⃑𝑗⋅⃑𝑖+4βƒ‘π‘—ο†βˆ’2=π‘š+8π‘š=βˆ’10.

So, the value of π‘š is βˆ’10, and ⃑𝐹=ο€Ίβˆ’10⃑𝑖+2⃑𝑗N.

Finally, we consider the case of a body subjected to a conservative force and another force acting against the conservative force so that the potential energy increases, for instance, when an object is lifted by a crane or when someone compresses or expands a spring. If we assume that the body started moving from rest and reached its final position with zero velocity, there is no change in the object’s kinetic energy between the two positions. According to the work–energy principle, we know that the sum of the works of all forces acting on the body is equal to the change in the kinetic energy. Hence, we have here that π‘Š+π‘Š=0,FcFo where π‘ŠFc denotes the work of the conservative force and π‘ŠFo the work of the other force. Since π‘Š=βˆ’Ξ”πΈFcP, we conclude that the work of the other force is equal to the change in potential energy: π‘Š=Δ𝐸.FoP

So, when a body is lifted by a force, this force does a positive work equal to the increase in its potential energy, while the weight does a negative work.

Key Points

  • The gravitational potential energy, 𝐸G, of an object with a mass π‘š that is a height β„Ž above the ground is given by 𝐸=π‘šπ‘”β„Ž,G where 𝑔 is the local acceleration due to gravity with a value of approximately 9.8 m/s2.
  • For a conservative force, the change in the potential energy of an object, Δ𝐸P, is equal to the negative of the work done, π‘Š, by the force on the object: Δ𝐸=βˆ’π‘Š.P

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.