Lesson Explainer: The Workโ€“Energy Principle Mathematics

In this explainer, we will learn how to use the workโ€“energy principle to solve problems of motion of a particle.

Let us first recall the definition of kinetic energy.

Definition: Kinetic Energy

The kinetic energy of a particle is dependent on the mass and velocity of the particle according to the formula ๐ธ=12๐‘š๐‘ฃ,K๏Šจ where ๐‘š is the mass of the body and ๐‘ฃ is the velocity of the body.

Work can be done on a particle by applying a force on it.

Definition: Work Done on a Particle by a Constant Force

The work done on a particle by a constant force to move the particle from an initial position to a final position is given by the scalar product between the force and the displacement vector between both positions: ๐‘Š=โƒ‘๐นโ‹…โƒ‘๐‘ .

Knowing the properties of the scalar product, if โƒ‘๐น is constant over a given displacement, then we have ๐‘Š=๐น๐‘ ๐œƒ,cos where ๐น is the magnitude of the force, โƒ‘๐น, ๐‘  is the magnitude of the displacement of the particle, โƒ‘๐‘ , and ๐œƒ is the angle between โƒ‘๐น and โƒ‘๐‘ . In this explainer, we will only deal with situations where โƒ‘๐น and โƒ‘๐‘  are parallel, so that ๐‘Š=ยฑ๐น๐‘ , depending on whether โƒ‘๐น and โƒ‘๐‘  have the same or opposite directions. A force whose work is negative is called a resistive force because it acts against the motion of the particle or body.

If โƒ‘๐น varies during the motion of the particle, we need to use the proper mathematic definition of the work done by a force during the motion from point ๐ด to point ๐ต: ๐‘Š=๏„ธโƒ‘๐นโ‹…โƒ‘๐‘ .๏Œก๏Œ d

We see that if โƒ‘๐นโ€™s orientation varies along the path between ๐ด and ๐ต, as it is the case for friction along a curved path, the work done by โƒ‘๐น depends on the path taken. By contrast, for weight, the work done depends on the initial and end points ๐ด and ๐ต only.

A force is said to be conservative if the total work done in moving the particle between two points does not depend on the path taken. It follows that if a particle comes back to its initial position, the total work done by this force is zero. Weight is a conservative force, while friction is a nonconservative force.

The workโ€“energy principle links the net work done on a body and the change in the kinetic energy of the body.

Definition: The Workโ€“Energy Principle

The change in the kinetic energy of a body is the sum of the work done on the body by the forces acting on it, regardless of whether or not these forces are conservative forces: ๐‘Š=๏„š๐‘Š=ฮ”๐ธ.netallforcesK

Let us see with our first example how the workโ€“energy principle enables us to determine the force that produces a change in the velocity of a body in an indeterminate time interval, only requiring us to know the change in velocity of the body and the displacement of the body.

Example 1: Using the Workโ€“Energy Principle to Find the Magnitude of a Force

A body of mass 96 kg was moving in a straight line at 17 m/s. A force started acting on it in the opposite direction to its motion. As a result, over the next 96 m, its speed decreased to 11 m/s. Using the workโ€“energy principle, determine the magnitude of the force.

Answer

According to the workโ€“energy principle, we have ๐‘Š=ฮ”๐ธ=๐ธโˆ’๐ธ=12๐‘š๏€น๐‘ฃโˆ’๐‘ฃ๏…,netKKKfifi๏Šจ๏Šจ where ๐ธKf and ๐ธKi are the final and initial kinetic energies, respectively, and ๐‘ฃf and ๐‘ฃi are the final and initial speeds. The mass of the body is 96 kg, and its initial and final speeds are17 m/s and 11 m/s respectively. Hence, ฮ”๐ธ=12ร—96ร—๏€น11โˆ’17๏…ฮ”๐ธ=โˆ’8064.KK๏Šจ๏ŠจJ

As ๐‘Š=ฮ”๐ธnetK, the work done by the force on the body is โˆ’8064 joules. We have ๐‘Š=๐น๐‘ ๐œƒ,cos where ๐น is the magnitude of the force, ๐‘  is the magnitude of the displacement of the particle, and ๐œƒ is the angle between โƒ‘๐น and โƒ‘๐‘ . The force acts on the body in the opposite direction to its motion. Hence, ๐œƒ=180โˆ˜, and so ๐‘Š=โˆ’๐น๐‘ .

This work is done over a distance of 96 m. The body moves in a straight line; therefore, the distance traveled equals the displacement, ๐‘ . Substituting in these values, we get โˆ’8064=โˆ’96๐น.

Rearranging to make ๐น the subject gives ๐น=โˆ’โˆ’8064โˆ’96=84.N

The magnitude of the force is 84 N.

Now let us look at another example involving resistive forces.

Example 2: Using the Workโ€“Energy Principle to Calculate the Ratio of the Resistance

Two bullets of equal mass were fired toward a target at the same speed but in opposite directions. The target was formed of two different pieces of metal stuck together. The first was 9 cm thick, and the second was 12 cm thick. When the bullets hit the target, the first one passed through the first layer and embedded 4 cm into the second before it stopped, whereas the other bullet passed through the second layer and embedded 5 cm into the first layer before it stopped. Using the workโ€“energy principle, calculate the ratio of the resistance of the first metallic layer to that of the second.

Answer

The question states that the bullets have the same mass and the same speed as each other, and so they have the same kinetic energy as each other. As the bullets have equal kinetic energy, equal work must be done on each bullet to bring it to rest.

For each bullet, two different resistive forces act on the bullet, corresponding to the resistances of the layers that they travel in. For each bullet, the resistive forces that act on it act for different distances, corresponding to the distances moved through the layers.

For the first bullet, the resistive force due to moving through the first layer acts over the entire thickness of the layer, 9 cm. The resistive force due to moving through the second layer acts over a distance of 4 cm. The paths of the bullets through the layers is shown in the following figure.

Writing the magnitude of the resistive forces due to the first and second layers ๐น๏Šง and ๐น๏Šจ, respectively, as these forces act parallel but in opposite direction to the motion, the work done on the first bullet, ๐‘Š, is given by ๐‘Š=โˆ’9๐นโˆ’4๐น.๏Šง๏Šจ

The second bullet moves through the entire thickness of the second layer and moves part of the way through the first layer. Treating the second bullet in the same way as the first bullet, we obtain ๐‘Š=โˆ’5๐นโˆ’12๐น.๏Šง๏Šจ

The work done on either bullet is equal to the work done on the other bullet, so โˆ’9๐นโˆ’4๐น=โˆ’5๐นโˆ’12๐น.๏Šง๏Šจ๏Šง๏Šจ

Multiplying each side by โˆ’1, we get 9๐น+4๐น=5๐น+12๐น.๏Šง๏Šจ๏Šง๏Šจ

This can be rearranged to determine the ratio of the two forces: 9๐นโˆ’5๐น=12๐นโˆ’4๐น4๐น=8๐น๐น=2๐น.๏Šง๏Šง๏Šจ๏Šจ๏Šง๏Šจ๏Šง๏Šจ

Therefore, the ratio of ๐น๏Šง to ๐น๏Šจ is 2โˆถ1.

Let us consider an example of how the workโ€“energy principle can be used to work out the kinetic energy of a body projected downward when it is about to hit the ground.

Example 3: Finding the Kinetic Energy of a Body Projected Downward When it is about to Hit the Ground

A body of mass 400 g was projected at 4 m/s vertically downward from a point 5 m above the ground. Use the workโ€“energy principle to calculate the bodyโ€™s kinetic energy when it was about to hit the ground. Take ๐‘”=9.8/ms๏Šจ.

Answer

According to the workโ€“energy principle, the work of all forces acting on the body is equal to the change in the kinetic energy of the body: ๐‘Š=ฮ”๐ธ=๐ธโˆ’๐ธ.netKKKfi

We are looking for ๐ธKf, so we need to work out ๐ธKi and ๐‘Šnet.

As the body is projected, it has an initial kinetic energy. In determining the initial kinetic energy of the body, its mass is converted from grams to kilograms: ๐ธ=12๐‘š๐‘ฃ๐ธ=12ร—0.4ร—4=3.2.KKii๏Šจ๏ŠจJ

The only force acting on the body is its weight. The work of a force is given by ๐‘Š=๐น๐‘ ๐œƒ,cos where ๐น is the magnitude of the force, โƒ‘๐น, ๐‘  is the magnitude of the displacement of the particle, โƒ‘๐‘ , and ๐œƒ is the angle between โƒ‘๐น and โƒ‘๐‘ . As the weight is parallel and in the same direction as the bodyโ€™s displacement, its work from the position 5 m above the ground to the ground is ๐‘Š=5๐‘š๐‘”=5ร—0.4ร—9.8=19.6.J

Applying now the workโ€“energy principle, we find ๐‘Š=๐ธโˆ’๐ธ19.6=๐ธโˆ’3.2๐ธ=3.2+19.6=22.8.netKKKKfifJ

The bodyโ€™s kinetic energy when it was about to hit the ground is 22.8 joules.

Let us now consider an example where the workโ€“energy principle is used to determine the resistive force that acts on a body.

Example 4: Using the Workโ€“Energy Principle to Calculate the Resistance

A body of mass 125 kg fell vertically from a height of 112 cm onto a section of sand. It sank 5 cm into the sand before it came to rest. Using the workโ€“energy principle, calculate the resistance of the sand to the bodyโ€™s motion. Take ๐‘”=9.8/ms๏Šจ.

Answer

According to the workโ€“energy principle, the work of all forces acting on the body is equal to the change in the kinetic energy of the body: ๐‘Š=ฮ”๐ธ=๐ธโˆ’๐ธ.netKKKfi

The body is described as falling from a height, so it is taken that it has no kinetic energy before it falls, that is, ๐ธ=0Ki. The body comes to rest at the end, which means that ๐ธ=0Kf. We thus have ๐‘Š=0net.

We now need to determine all the forces acting on the body; these are its weight and the force exerted bythe sand. Air resistance is neglected.

Let us calculate the work of each of these two forces. Recall that the work of a force is ๐‘Š=๐น๐‘ ๐œƒ,cos where ๐น is the magnitude of the force, โƒ‘๐น, ๐‘  is the magnitude of the displacement of the particle, โƒ‘๐‘ , and ๐œƒ is the angle between โƒ‘๐น and โƒ‘๐‘ . Both forces are acting parallel to the bodyโ€™s motion, but while the weight acts in the same direction as the bodyโ€™s motion, the sand resistance acts in the opposite direction (it is a resistive force indeed).

Hence, ๐‘Š=๐‘š๐‘”๐‘ ,weight where ๐‘  is the magnitude of the total displacement of the body as it descends 112 cm to reach the ground and a further 5 cm through the sand, as shown in the following figure.

The distance that the body moves is the 112 cm of the height of the body above the ground plus the 5 cm that the body penetrates the sand, for a total distance of 117 cm.

The 117 cm distance is converted to 1.17 m so that the units of distance are consistent with the units of metres per second squared used for the given value of ๐‘”. So, we have ๐‘Š=125ร—9.8ร—1.17=1433.25.weightjoules

The sand resistance acts only when the body penetrates the sand, that is, over the last 5 cm, or 0.05 m. Therefore, its work is ๐‘Š=โˆ’0.05๐น,SRSR where ๐นSR is the magnitude of the force exerted by the sand.

Applying now the workโ€“energy principle, we find ๐‘Š=๐‘Š+๐‘Š=01433.25โˆ’0.05๐น=00.05๐น=1433.25๐น=1433.250.05=28665.netweightSRSRSRSRN

The magnitude of the resistive force exerted by the sand is 28โ€Žโ€‰โ€Ž665 N.

Key Points

  • A force is said to be conservative, such as weight, if the total work done in moving a particle between two points does not depend on the path taken.
  • A force is said to be nonconservative, such as friction, if the total work done in moving a particle between two points depends on the path taken.
  • The workโ€“energy principle states that the change in the kinetic energy of a body is the sum of the work done on the body by all forces acting on it, regardless of whether or not these forces are conservative forces: ๐‘Š=๏„š๐‘Š=ฮ”๐ธ.netallforcesK

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