# Lesson Video: The Work–Energy Principle Mathematics

In this video, we will learn how to use the work–energy principle to solve problems of motion of a particle.

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### Video Transcript

In this video, we will learn how to use the work–energy principle to solve problems of motion of a particle. First, we must begin by defining what the work–energy principle is.

The work–energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. In equation form, we can say that 𝑊, the net work, is equal to ΔKE, which represents the change in kinetic energy. Recalling that kinetic energy is equal to one-half 𝑚𝑣 squared, where 𝑚 is the mass and 𝑣 is the speed of the object, we can expand the change in kinetic energy out to be one-half 𝑚𝑣 final squared minus one-half 𝑚𝑣 initial squared. And recalling that work is equal to 𝐹𝑑, where 𝐹 is the force that’s parallel to the displacement and 𝑑 is the displacement, we get an expanded equation for our work–energy principle.

We’re on top of solving for work or change in kinetic energy. We can also solve for force, displacement, mass, and initial and final speeds. This principle could be applied to problems to solve the motion of a particle, such as problems where a box comes to rest over a rough surface. Let’s apply the work–energy principle to a few examples, starting with finding the kinetic energy of a body projected downward when it’s about to hit the ground.

A body of mass 400 grams was projected at four meters per second vertically downward from a point five meters above the ground. Use the work–energy principle to calculate the body’s kinetic energy when it was about to hit the ground. Take 𝑔 equals 9.8 meters per second squared.

We can draw a diagram with the labeled information from the problem about our object. It has a mass of 400 grams. It’s thrown vertically downward at a speed of four meters per second from a height of five meters. And we are looking to find the kinetic energy when the object is just about to hit the ground. We are instructed to use the work–energy principle. In equation form, this means that the net work done on an object is equal to the change in kinetic energy of the object. The net work done on an object is equal to the net force on the object times the displacement. So we can replace 𝑊 with 𝐹 times 𝑑. Change in kinetic energy means the final kinetic energy minus the initial kinetic energy. We should recall that the kinetic energy of an object is equal to one-half 𝑚𝑣 squared, where 𝑚 is the mass of the object measured in kilograms and 𝑣 is the speed of the object measured in meters per second.

As we’re solving for the final kinetic energy, we don’t need to change this variable. However, the problem did not give us the initial kinetic energy, so we need to replace this variable with an equal form, which would be one-half 𝑚𝑣 initial squared. The net force 𝐹 acting on the object is the force due to gravity as it’s pulling the object to the ground. We should remember that the force due to gravity is the mass of the object measured in kilograms times the acceleration due to gravity, which was given to us as 9.8 meters per second squared in the problem. We can go ahead and replace the net force in our equation with 𝑚𝑔. If we add the initial kinetic energy, the one-half 𝑚𝑣 initial squared, to both sides, we will now have an expression for the final kinetic energy. The final kinetic energy of the object is equal to 𝑚𝑔𝑑 plus one-half 𝑚𝑣 initial squared.

Before we plug in any values from our problem, we have to first make sure that they’re all in the correct units. Looking over our values, the only variable that needs to be converted is the mass, from grams to kilograms. One kilogram is equal to 1,000 grams. When we divide 400 grams by 1,000, we end up with a mass for our object of 0.400 kilograms. We replace the mass of the object with 0.400, 𝑔 with 9.8, 𝑑 with five, and the initial speed with four. When we multiply out our first term, we get 19.6. And when we multiply out our second term, we get 3.2. Adding these two terms together, we end up with a final kinetic energy of 22.8 joules, where joules is the standard unit for energy and is represented by the capital J. Using the work–energy principle, the body’s final kinetic energy just as it’s about to hit the ground is 22.8 joules.

Our next example problem will use the work–energy principle to find the magnitude of a force.

A body of mass 96 kilograms was moving in a straight line at 17 meters per second. A force started acting on it in the opposite direction to its motion. As a result, over the next 96 meters, its speed decreased to 11 meters per second. Using the work–energy principle, determine the magnitude of the force.

In the problem, we are instructed to use the work–energy principle. In equation form, this states that the net work down on an object is equal to the change in kinetic energy of the object. Recalling that work is defined as force times displacement, where the force is parallel to the displacement, we can expand out our formula, replacing 𝑊 with 𝐹 times 𝑑. Change in kinetic energy is the final kinetic energy minus the initial kinetic energy. We should remember that the kinetic energy of an object is equal to one-half 𝑚𝑣 squared, where 𝑚 is the mass of the object and 𝑣 is the speed of the object. We can replace kinetic energy final with one-half 𝑚𝑣 final squared and kinetic energy initial with one-half 𝑚𝑣 initial squared. To isolate the force, we can divide both sides of the equation by 𝑑, which will cancel out the displacement on the left side.

Now that we have an expression for our force, we can substitute in the values from our problem. We use 96 for the mass, 11 for the final velocity, 17 for the initial velocity, and 96 for the displacement. When we multiply out our first term, one-half times 96 times 11 squared, we get 5,808. Multiplying out the second term of one-half times 96 times 17 squared, we get 13,872. When we subtract our numerator and divide by our denominator, we get a force of negative 84 newtons. We are asked to find the magnitude of the force, and therefore we do not need the negative sign as that tells us the direction. Using the work–energy principle, the magnitude of the force acting on the object is 84 newtons.

In the next example, we will be finding the loss in kinetic energy after the collision of two balls moving in opposite directions.

Two spheres were moving in opposite directions along a horizontal line. The first sphere had a mass of six kilograms, and its speed was 75 centimeters per second when it collided with the second sphere, which was moving at 80 centimeters per second. As a result of the impact, the first sphere rebounded at 15 centimeters per second along the same line in the opposite direction, and the second sphere came to rest. Find the loss in kinetic energy as a result of the impact.

We can draw a diagram of the problem below. Sphere one will be represented by the blue circle, and sphere two will be represented by the pink circle. Before the collision, sphere one has a mass of six kilograms and it’s traveling at 75 centimeters per second. After the collision, the same sphere is now traveling at 15 centimeters per second in the opposite direction. Before the collision, sphere two is traveling toward sphere one at a speed of 80 centimeters per second. After the collision, sphere two comes to rest, which means it has a speed of zero centimeters per second. We’re being asked to find the loss in kinetic energy, remembering that the kinetic energy of an object is equal to one-half 𝑚𝑣 squared, where 𝑚 is the mass and 𝑣 is the speed of the object. We need the mass of both objects to determine the loss in kinetic energy. However, we do not have the mass of sphere two.

We can use conservation of momentum to determine the unknown mass. Remember that conservation of momentum states that the initial momentum of a system equals the final momentum of a system. Recalling that the momentum of an object is equal to the mass of the object times the object’s velocity, we can expand out our conservation of momentum equation. When we plug in the values for the momentum of each sphere before the collision and after the collision, we end up with the expression below. Before we can do any calculations, we do need to make a unit conversion. Our speeds were given to us in centimeters per second but need to be converted into meters per second. One meter is equal to 100 centimeters, so we can divide each of our velocities by 100 to convert them from centimeters per second to meters per second. We now have 0.75, 0.80, negative 0.15, and zero for our velocities.

Now we can multiply out each of the terms in our equation. We get 4.50 minus 0.80 times 𝑚 is equal to negative 0.90 plus zero. To isolate 𝑚, we can add 0.80𝑚 to both sides of the equation and 0.90 to both sides of the equation. Our final step to find the mass would be to divide both sides by 0.80, which gives us a mass for sphere two of 6.75 kilograms. Now we can determine our loss in kinetic energy. To find the loss in kinetic energy, we must subtract the initial kinetic energy from the final kinetic energy. Using our equation for kinetic energy, we can expand out our loss in kinetic energy equation. After the collision, only sphere one has kinetic energy as sphere two comes to rest. Therefore, we only have one object that has a final kinetic energy, which can be expressed as one-half times six times 0.15 squared.

We have two initial kinetic energies as both sphere one and sphere two are moving before the collision. Notice that we once again converted our speeds to meters per second before we plug them in to the equation. Multiplying out our variables, we get 0.0675 for our first term, 1.6875 for our second term, and 2.16 for our third term. When we subtract our initial kinetic energies from our final kinetic energy, we end up with a loss of negative 3.78 joules, where the negative symbol represents that we have lost energy. The loss in kinetic energy as a result of the impact is 3.78 joules.

In the next example problem, we will use the work–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 nine centimeters thick, and the second was 12 centimeters thick. When the bullets hit the target, the first one passed through the first layer and embedded four centimeters into the second layer before it stopped, whereas the other bullet passed through the second layer and embedded five centimeters into the first layer before it stopped. Using the work–energy principle, calculate the ratio of the resistance of the first metallic layer to that of the second.

The diagram below represents a visualization of the problem, with the yellow box being the first layer with a thickness of nine centimeters and the pink box representing the second layer with a thickness of 12 centimeters. The top blue dotted line represents the first bullet, which ended up being embedded four centimeters into the second layer. And the bottom blue dotted line represents the second bullet, which ended up being embedded five centimeters into the first layer.

The problem tells us to use the work–energy principle. In equation form, that is that the net work done on an object is equal to the change in kinetic energy of the object. We can apply this equation to both bullets. Recall that the net work done on an object is equal to the force times the displacement. The net work done on bullet one is 𝐹 one times nine plus 𝐹 two times four. 𝐹 one represents the resistance of layer one and nine is the thickness as the bullet goes completely through layer one. And 𝐹 two represents the resistive force of layer two and four for the distance as the bullet is embedded four centimeters into layer two. We can leave the change in kinetic energy as ΔKE, as both bullets have the same mass, the same initial speed, and eventually came to rest.

The expanded equation for the second bullet is 𝐹 one times five plus 𝐹 two times 12 equals ΔKE. 𝐹 one and 𝐹 two once again represent the resistive force of the respective layers, layer one and layer two. The five is the distance that the second bullet gets embedded into the first layer, five centimeters. And the 12 comes from the fact that the bullet goes all the way through the second layer, which is 12 centimeters thick. We wanna find the ratio of 𝐹 one to 𝐹 two or the resistance of the first metallic layer to that of the second. To do this, we can subtract our equations. 𝐹 one times nine minus 𝐹 one times five is 𝐹 one times four. And 𝐹 two times four minus 𝐹 two times 12 is negative 𝐹 two times eight. When we subtract our two change in kinetic energies, we get zero, as the bullets are the same mass, same initial speed, and come to rest.

Our next step is to add 𝐹 two times eight to both sides of the equation. To rearrange our equation so that we have a ratio, we can divide both sides by 𝐹 two and both sides by four. The ratio of 𝐹 one to 𝐹 two is two to one, more commonly written in this form. The ratio of the resistance of the first metallic layer to that of the second is two to one.

Now that we’ve discussed the work–energy principle and done four example problems, let’s go over the key points of this lesson.

Key Points

Net work on an object is equal to the change in kinetic energy of the object. In equation form, that would be 𝑊 equals ΔKE, which is also known as the work–energy principle. The work–energy principle can be used to solve problems involving a moving particle under the action of a constant force.