Physics Explained Through a Video Game/Introduction to Work and Energy

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Topic 3.1 - Introduction to Work and Energy

Goals:

• Know what work is.
• Understand and apply the derivation for work.
• Know what kinetic and potential energy are.
• Use and calculate physical situations using the work-energy theorem.

Topic 3.1.1 - Work

Example 1: Rolling City Boulder

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To further describe the behavior of a physical object, we can consider work. Work is a type of energy within a system, which we'll discuss shortly. It involves considering the magnitude of a selected force on an object and the displacement that the object has had.

In its simplest form, work is the product of a force's magnitude and an object displacement, given that (1) the force is constant and (2) the force and displacement vectors are pointing in the same direction. In this special case, $\text{Work}=\text{Force}\cdot \text{Displacement}$. We could use the convention that work can be represented by the symbol $W$ such that $W=\overrightarrow{F}\overrightarrow{d}$.

As an example of this, consider a rolling circular boulder in a modified version of lack of comfort by swiped3. The boulder accelerates towards the right while on a flat surface. This is because wind is pushing on the boulder, providing it a constant ${\overrightarrow{F}}_{\text{net}}$ that is pointed towards the right. Additionally, during the video, there is a displacement vector towards the right.

Exercise: Suppose that ${\overrightarrow{F}}_{\text{net}}$ has a magnitude of $10.\text{N}$ and the displacement magnitude is $5.0\text{m}$. What would be the net work done on the boulder?

Answer:

In this case, we are told that the force is constant. Also, we can see in the diagram that the ${\overrightarrow{F}}_{\text{net}}$ and $\overrightarrow{d}$ vectors are pointing in the same direction (towards the right). Thus, we can use the formula $W=\overrightarrow{F}\overrightarrow{d}$, as shown below to find the work done on the boulder by the net force, ${\overrightarrow{F}}_{\text{net}}$.

$W=\overrightarrow{F}\overrightarrow{d}$

$W_{\text{net}}={\overrightarrow{F}}_{\text{net}}\overrightarrow{d}$ [We're considering the work done by the net force on the ball. As such, we're calculating for the net work on the ball.]

$W_{\text{net}}=(10.)(5.0)\text{N}\cdot \text{m}$ [Substitution.]

$W_{\text{net}}=50.\text{N m}$ [2 sigfigs]

The boulder has a net work of $50.\text{N m}$ while it is pictured rolling on the rooftop.

Example 2: Propeller Planes

File:PlaneWork.webmAs previously seen in Topic 2.3 - Newton's Second Law, let's again consider the map Maneuver Planes by TiM MELL0 where the player Sam (red) descends a plane towards the ground. We'll use this example to generalize the concept of calculating the work for a constant force on an object.

To note, we can also calculate for the work done by a force on an object even when the displacement vector isn't parallel with the force vector. With this, we can use the general formula for the work done on an object, given a constant force:

Definition of Work Given a Constant Force:

$W=\overrightarrow{F}\mathit{\cdot }\overrightarrow{d}$

With the equation above, note two things:

1. Work is a scalar value. This means that it is a one-dimensional value. However, it can be either negative or positive. For more information, consider this video by Khan Academy. Also, the next example will go further into this concept.
2. The dot product is used in the equation. This is a type of multiplication between two vectors. Generally, this concept is introduced in a precalculus course.

If dot products are unfamiliar to you, we can also write the general equation for work done by constant force alternatively as $W=Fd\cos (\theta )$. In this,

• $F$ and $d$ are respectively the magnitude of the vectors $\overrightarrow{F}$ and $\overrightarrow{d}$.
• $\theta$ is the difference in direction between the two vectors.

A diagram calculating for the difference between the direction angles of the gravitational force and displacement vectors on a plane.

Exercise: Suppose that there's a gravitational force of $8.0\cdot 10^4\text{N}$ (directed downwards) on the plane as it is descending. Using two points on the plane's path, labeled as "START" and "END," there is found to be a distance of $100.\text{m}$ between them. Additionally, the displacement vector of the plane between these two points is directed $30{.}^{\circ }$ below the +X Axis. What is the work done by the gravitational force between the two labeled points?

Answer: To solve this problem, we can first list the information mentioned in the prompt.

• We're solving for the work done by the gravitational force on the plane.
• The gravitational force vector has a magnitude of $8.0\cdot 10^4\text{N}$ and is directed downwards.
• The displacement vector has a magnitude of $100.\text{m}$. It is directed $30{.}^{\circ }$ below the +X Axis.

With this, we can find the variables needed for substitution into the equation $W=Fd\cos (\theta )$. More specifically, the gravitational force vector's magnitude is our $\overrightarrow{F}$ variable. In addition, the displacement vector represents the $\overrightarrow{d}$ variable. Therefore, we can partially substitute in our variables into the equation such that:

$W=Fd\cos (\theta )$ [Formula]

$W_{grav}=F_{grav}d\cos (\theta )$ [We're considering the gravitational force acting on the plane.]

$W_{grav}=(8.0\cdot 10^4)(100.)\cos (\theta )\text{N m}$ [Substitution of the gravitational force and displacement vector magnitudes.]

To consider the value of $\theta$, we can graph ${\overrightarrow{F}}_{grav}$ and $\overrightarrow{d}$ as diagrammed to the right. Using the information given and the diagram, we can find that $\theta =60{.}^{\circ }$. We can substitute this value into our equation for $W_{grav}$ and continue simplifying.

$W_{grav}=(8.0\cdot 10^4)(100.)\cos (60{.}^{\circ })\text{N m}$ [Substitution of $\theta$]

$W_{grav}=4.0\cdot 10^6\text{N}\text{m}$ [Using degree mode; Algebra; 2 sigfigs]

While the plane is descending, the gravitational force does $4.0\cdot 10^6\text{N}\text{m}$ of work on the plane.

Example 3: Assembly Line

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File:CopyMapFactory.webm We can also calculate the work done by a force even when a force is changing. Suppose that in the map Factory of Copy Maps by antidonaldtrump / Armin van Buren that colored boxes are being pushed along an assembly line. On the top row of the assembly line, the blocks have a net force that is pushing them forward along the line, accelerating them in the leftward direction.

Unlike previous examples, the force that is being exerted on the object is varying. This means that the equation $W=Fd\cos (\theta )$, which assumes that the force is constant, cannot be directly used.

Instead, we can consider calculating for the work on the colored boxes graphically.

Walkthrough: Suppose that on the top row of the assembly line where:

• The yellow box flips onto it side and begins moving towards the left over two conveyor belts.
• The yellow box travels $1.0\text{m}$ along the first belt.

• The net force for the yellow box on the first belt is $200.\text{N}$.
• Then, the yellow box travels $4.0\text{m}$ along the second belt.
• The net force for the yellow box while on the second belt is $300.\text{N}$.

In this example, we're going to be calculating graphically for the work done on the yellow box by the net force. To do this, we can create a function of the force on the yellow box with respect to its displacement as pictured on the right. To go further, we need to consider some theory regarding work.

The work done by a force is the accumulation of that force over a distance. In other words, we can graph a function of force with respect to distance. Since work is the accumulation of force with respect to distance, the area under the above-mentioned function is the work.

A graph of the force on a yellow box as it is pulled down an assembly line in a factory.

Using this definition, we can create a graph of the yellow box's net force, ${\overrightarrow{F}}_{net}$ with respect to $\overrightarrow{d}$. Because the force exerted on the yellow box is dependent on where the box is positioned, ${\overrightarrow{F}}_{net}$ is the independent variable ($Y\text{axis}$) and $\overrightarrow{d}$ is the dependent variable ($X\text{axis}$).

As mentioned above, because work is the accumulation of force with respect to distance, if we were to calculate the area between the green curve and the $X$ axis, this would be work done on the yellow box between $0.0\text{m}$ and $5.0\text{m}$.

This can be done through breaking apart the pictured graph into two rectangles as also pictured. From here, we can easily calculate the area of each of these rectangles, giving us the amount of work done.

With this, we can combine the two rectangle areas to find the work done:

Shape	Shape Area	Associated Displacement Domain
Purple Rectangle	$200.\text{Nm}$	$0.0$ to $1.0\text{m}$
Blue Rectangle	$1200.\text{Nm}$	$1.0$ to $5.0\text{m}$
Total Area	$1400.\text{Nm}$ ($1.40\cdot 10^3\text{Nm}$ when adjusted for sigfigs)	$0.0$ to $5.0\text{m}$

Thus, when the yellow box was being pushed along the top portion of the assembly line, it had a work of $1.40\cdot 10^3\text{Nm}$ done by the net force acting on it.

Section 3.1.2 - Kinetic Energy

Example 1: Arrows in the Forest

File:ForestArrows.webm In this example, there will be an introduction of kinetic energy, what it means conceptually, and how to calculate for it in a physical situation. [Content]

Example 2: Flying Squirrel

File:FlyingSquirrel.webm

[Intro] Suppose the in-game situation in the map Flying Squirrel by Raspy 667. In this map, a flying squirrel is gently moving through the air back and forth. What would be the kinetic energy of the squirrel at each of the labeled points from the video [cont]

Consider discussing your solutions on this article's Talk Page. On there, you can find help from others. Template:Navigation

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