# Physics Explained Through a Video Game/Introduction to Work
### Topic 3.1 - Introduction to Work

Goals:
* Know what work is.
* Understand and apply the derivation for work.
* Understand how work and force are related concepts graphically.

#### Example 1: Rolling City Boulder

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 $W={\vec {F}}{\vec {d}}$.
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 ${\vec {F}}_{\text{net}}$ that is pointed towards the right. Additionally, during the video, there is a displacement vector towards the right.
Exercise: Suppose that ${\vec {F}}_{\text{net}}$ 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 ${\vec {F}}_{\text{net}}$ and ${\vec {d}}$ vectors are pointing in the same direction (towards the right). Thus, we can use the formula $W={\vec {F}}{\vec {d}}$, as shown below to find the work done on the boulder by the net force, ${\vec {F}}_{\text{net}}$.
$W={\vec {F}}{\vec {d}}$
$W_{\text{net}}={\vec {F}}_{\text{net}}{\vec {d}}$ [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

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: | $W={\vec {F}}{\boldsymbol {\cdot }}{\vec {d}}$ |
With the equation above, note two things:
* Work 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.
* 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.
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 ${\vec {F}}$ and ${\vec {d}}$.
* $\theta$ is the difference in direction between the two vectors.
$8.0\cdot 10^{4}\;{\text{N}}$
$100.\;{\text{m}}$
$30.^{\circ }$
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 ${\vec {F}}$ variable. In addition, the displacement vector represents the ${\vec {d}}$ 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 ${\vec {F}}_{grav}$ and ${\vec {d}}$ 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

Content Prerequisite Before reading this example, please watch a related video created by Khan Academy.
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.
Using this definition, we can create a graph of the yellow box's net force, ${\vec {F}}_{net}$ with respect to ${\vec {d}}$. Because the force exerted on the yellow box is dependent on where the box is positioned, ${\vec {F}}_{net}$ is the independent variable ($Y\;{\text{axis}}$) and ${\vec {d}}$ 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.

#### Question 1: Colorful Canyon

In the diagram, as labeled by a solid green path, the player briefly slides across a portion of the rock face. Then, labeled by a solid yellow path, the player slides off the rock face, entering free fall motion until striking a boulder in the river. Dotted green and yellow lines are provided, indicating the magnitude of displacement in both the horizontal and vertical directions for both parts of the diagrammed path.
Assume that:
* The rockface has negligible friction on the player.
* The rockface area underneath the green path has a length of $1.25\;{\text{m}}$.
* The player has a mass of $40.\;{\text{kg}}$.
* The acceleration due to gravity on the player is $9.8\;{\frac {\text{m}}{{\text{s}}^{2}}}$.
   Consider when the player is sliding across the rock face portion under the green path.
   (i) Calculate the magnitude of the gravitational force acting on the player.
   (ii) Calculate the normal force that is acting on the player.
   (iii) Draw a completed free body diagram. Use the unlabeled reference image on the right.
   Continue to consider when the player is sliding across the rock face portion under the green path.
   (i) Determine the net force magnitude and direction of the player.
   (ii) Calculate the work done by the net force on the player when traveling along the green path.
   Now consider when the player is traveling in free fall.
   (i) Determine the net force magnitude and direction of the player.
   (ii) Calculate the overall displacement of the player while on the yellow portion of the path.
   (iii) Calculate the work done by the net force on the player when in free fall.
   Now consider the player over the entire labeled path.
   (i) Create a graph of the net force acting on the player with respect to the total displacement of the player.
   (ii) Calculate the work done by the net force on the player when traveling along the green path.
Consider discussing your solutions on this article's Talk Page. On there, you can find help from others.