Larger vehicles require more energy to move than smaller vehicles Image Credit: iStock.com/DarthArt

Work and Energy

In previous modules, you learned how Newton’s Laws can be used to solve a wide variety of problems and how vectors are a valuable tool for analyzing motion in two dimensions. You will now learn another way of solving problems that is complementary to dynamics and kinematics.

Our earlier work examined particles and their changes in speed or direction that resulted from outside influences (forces) such as gravity. We will now explore energetics—the energy contained in a system. The ability to solve problems in energetics depends entirely on a fundamental law of physics called the Law of Conservation of Energy. This law states that the total energy in a system can never be created or destroyed. Well, that definition raises the question: What is a system? For our purposes at this time, you can think of a system as the set of objects that interact with each other to cause the observed action. This might not be the most complete definition, but it’s sufficient to get you started.

Let’s start by connecting what you just studied—force—with energy. You’re familiar with the concept of kinetic energy, otherwise known as the energy of motion. It takes more energy, in the form of gasoline, to get a large SUV to go from to 100 space km divided by hour than it does to get a tiny Smart Car to do the same. Why? Because the Smart Car is lighter, meaning less massive. Also, it takes more energy to get that tiny car to go from 0 to 100 km divided by hour than it takes to get it to go from 0 to 50 km divided by hour . That, too, should be obvious—of course it takes more gasoline to get it to go faster.

Now suppose that, instead of a gasoline-powered vehicle, it was human-powered, like a wooden go-cart. Ignore friction and think about how to get the go-cart to go fast and faster. If you push with a constant force, you can get the go-cart to go fast, but if you push with that same force over a longer distance, you can get it to go faster. Therefore, the longer you exert a constant force on the go-cart the more kinetic energy it will have. What if there is only a short distance to push? How could you get the go-cart to move faster in that case? You can have a friend help you push. In this case, having a larger force over a given distance will lead to more kinetic energy.

Watch the following video to learn more about work and energy.

Introduction to Work and Energy. Khan Academy, 2008. YouTube. Web. Accessed 20 April 2018.

Man pushes a child in a go kart as adults watch on the side of the road.

Applying a constant force to an object over a distance gives the object kinetic energy. | Image Credit: iStock.com/BluIz60 (Links to an external site.)

Force, Distance, and Kinetic Energy

These vehicle scenarios show a link between force, distance and kinetic energy. Let’s apply one more component to see the effect of another variable. When you imagined pushing the go-cart, for example, at what angle were you pushing? Were you standing above the cart pushing straight down, or were you standing behind the cart pushing as directly as possible in the direction of travel? You probably imagined yourself pushing the cart directly in the direction of travel. What is the best angle at which to apply the force? A 90° angle means pushing straight down, and you can see immediately that that will get the go-cart nowhere. A 0° angle means pushing directly in the direction of travel. You can see, from our everyday experience, that would be the optimum angle at which to apply the force.

Here is one final piece for you to consider: Energy is a scalar quantity. Putting all these pieces together, you can see that kinetic energy depends on the mass of an object, the amount of force applied while moving that object (without friction), and the displacement and angle over which that force is applied. We need a way to combine all these ingredients—including two vectors (force and displacement)—to get a scalar quantity: the kinetic energy.

Large cardboard box on a white background

Work is the energy obtained from a force applied to an object, such as a box, over a given distance. | Image Credit: iStock.com/kyoshino (Links to an external site.)

Scalar-from-Vector

Before getting to the scalar-from-vector trick, let’s define the basic concept that ties all these physical quantities together. Work is the energy obtained from a force applied to an object over a given distance. We will consider only the component of the force vector that is in the same direction as the displacement. So, if you’re pushing that go-cart at an angle of 30° relative to horizontal, then only the horizontal component of that force open parentheses cos open parentheses 30 degree close parentheses close parentheses goes into work, and the remaining vertical component of the applied force open parentheses sin open parentheses 30 degree close parentheses close parentheses goes into pushing the cart down into the ground, increasing the normal force but not helping you go any faster.

So, how do you get a scalar from two vectors? Use the scalar product. If you’ve taken Calculus 3 then you will recognize this as the dot product. The scalar product is simply the product of two vectors’ magnitudes and the cosine of the angle between them. That should make sense in our example. Pushing straight down yielded no work, even though there was a force, because cos open parentheses 90 degree close parentheses is zero. And pushing straight along the direction of travel gave us maximum kinetic energy (work) because cos open parentheses 0 degree close parentheses is 1.

We can now write our equation for work:

W space equals space F with rightwards arrow on top times increment r with rightwards arrow on top space equals space F increment r cos theta

where increment r with rightwards arrow on top is the change in position, or the displacement.

Finally, by what’s known as the work-kinetic energy theorem, we can find the change in kinetic energy of the object:

increment K E space equals space W

Let’s put these ideas into action. Calculate the final speed of a 15 kg box sliding on a horizontal frictionless surface after a 42 straight N force is applied at 27° relative to the horizontal, over a distance of 7.0 straight m .

First, we will use the work-kinetic energy theorem to calculate the final kinetic energy of the box:

W space equals space 42 N times 7.0 m times cos 27 degree space equals space 260 straight J

The work done on the box is 260 joules so the final kinetic energy of the box is 260 straight J :

K E space equals space 260 straight J space equals space 1 half m v squared

Given m space equals space 15 k g , solving for v yields a final speed of 5.9 straight m divided by straight s .

In the above examples, it was assumed that a constant force was being applied over a certain displacement. For example, the force of gravity decreases as you get farther from the center of the earth, and the restoring force of a spring, F space equals space minus k x , increases as displacement, x, increases. In these cases, calculus is needed to integrate the force over d x to get work. Please refer to your textbook to see the details of these calculations.

Summary

You may want to refer back to this lecture for support on the following concepts:

Law of Conservation of Energy
Work and energy
Connections among force, distance and kinetic energy
Scalar product of vectors