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===<big>Topic 2.3 - Newton's Second Law</big> ===

In order to apply the concepts of kinematics to a greater amount of physical situations, we need to consider how forces can be used to explain the movement of objects.
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====<big>Example 1: Rocket Propulsion</big>====
</div>[[File:ForceRocket.webm|thumb|right|335x335px|''<nowiki>*</nowiki>The map is a private map by monkey butler.'']]

As illustrated with a rocket*, objects accelerate in the direction of the net force applied onto them. In this case, the only non-negligible force acting on the rocket is its propulsion, which is pushing the rocket in the right-upwards direction. Thus, the rocket's propulsion force is its net force, <span><math>\vec F_{net} </math></span>.

With this, we can see that the rocket is accelerating from rest in the right-upwards direction. In other words, the rocket is speeding up in the right-upwards direction because of the propulsion force.

The behavior of the rocket's acceleration can be generalized by noting the definition of Newton's Second Law below:

{| align="center" style="border:solid 4px #888; padding:0.5em;"
 | valign="top" style="font-size:{{{quotesize|11pt}}}; padding:4px 5px;" |Definition of Newton's Second Law:
 | valign="top" style="font-size:{{{quotesize|11pt}}}; padding:2px 5px;" |At any instant, the net force acting on an object is equal to the object's mass multiplied by its net acceleration.<ref>“5.4: Newton’s Second Law.” ''Physics LibreTexts'', 18 Oct. 2016, <nowiki>https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/05%3A_Newton’s_Laws_of_Motion/5.04%3A_Newton’s_Second_Law</nowiki>.</ref>
 |
 |}
With this definition, we can create the equation <span><math>\vec F_{net} = m\vec a_{net} </math></span>. This allows for us to link together the concept of forces being applied onto objects and the object's acceleration.

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====<big>Example 2: Skateboard Riding</big>====
</div>Consider a modified version of ''Skate Duel ''by ''TiM MELL0''. In this, ''monkey butler'' (left) and ''Hello4409'' (right) both try to move their skateboards by applying a force that alternates between the left and right directions on a skateboard. To note, both players have a horizontally applied force of the same magnitude.

[[File:Skateboards.webm|center|620px]]

As shown by the video above, ''Hello4409's'' skateboard moves back and forth much more slowly than ''monkey butler's'' skateboard. This is because the skateboards have different masses. More specifically, ''Hello4409's'' skateboard has more mass. Thus, when ''Hello4409'' tries to apply a force onto his skateboard, it accelerates more slowly.
[[File:SkateboardDiagram(1).png|thumb|right|335x335px]]

This can be expressed by the diagram on the right, illustrating that given some known net force, <span><math>F_{net}</math></span>, if the object has more mass, then the net acceleration of that object, <span><math>a_{net}</math></span>, will be less.

In the case of the skateboard example, ''monkey butler'' and ''Hello4409's'' skateboards are undergoing the same forces when the skateboard is at the <u>bottom of the ramp</u>. More precisely, both player's skateboards have gravitational and contact forces that cancel out. As such, the net force for both players is the applied force. 

Although the applied forces are equal in magnitude, we can consider Newton's Second law where <span><math>F_{net} = ma_{net} \implies a_{net} = \frac{F_{net}}{m}</math></span>. Because ''monkey butler's'' skateboard (colored in blue) has less mass, it will have a greater magnitude of net acceleration.

Also, because ''monkey butler'' is able to give his skateboard a greater acceleration, his blue skateboard is able to travel further up the ramp. This concept will be further explored later on in Unit 3.

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====<big>Example 3: Car Chase</big>====
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For a further look into the applications of Newton's Second Law, consider a modified version of ''Car Chase ''by ''O_o O_o O_o'' below. In this map, a police car is pursuing a red vehicle where both vehicles gradually slow down and come to a stop.
[[File:CarChase.webm|thumb|center|620px]] <div>
'''Task: '''<br>Assume that <br> • Air resistance is negligible. <br> • The police car's gravitational force and contact force are found to be negligibly different in magnitude. <br> • The police car's applied force from engine thrust and the car's friction force from the rough ground are acting only horizontally.

Considering the video overall, which of the forces mentioned below has a greater magnitude while the police car is slowing down? Then, create a free body diagram (FBD) of a police car.
{| class="wikitable"
|-
|The police car's applied force (from engine thrust)
|The police car's friction force (from the rough ground)
| Neither. They are equal in magnitude.
|-
|}
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<span style="background:black; -webkit-background-clip:text !important; -webkit-text-fill-color:transparent;">'''''Answer:'''''</span><div class="mw-collapsible-content">

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The police car's friction force (from the rough ground).
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To explain, we already know that the police car is slowing down. Because the police car is initially moving towards the right and then enters rest relative to the ground, the police car has a net acceleration, <span><math> a_{net} </math></span> in the leftward direction. Using the [[Physics Explained Through a Video Game/Displacement|Cartesian Coordinate Convention]], the net acceleration is pointed in the negative horizontal direction.
[[File:PoliceCarFBDDiagram.png|thumb|right|335x335px]]
Because of Newton's Second Law, the direction of the net force is the same as the net acceleration. Thus, the police car has a net force that is pointing leftwards.

With the vertical forces, including the gravitational and contact forces are equal in magnitude, they fully cancel each other out. Also, we're aware that the friction force is acting towards the left and the applied force is acting the right. With this information, we can begin creating a free body diagram (FBD) of the situation, as shown on the right.

Since only the friction and applied forces are in the horizontal direction, for the net force to be in the negative (leftward) direction, the frictional force must have a '''greater magnitude'''. This is reflected by the FBD where the vector of the frictional force is drawn longer than the applied force.
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<span style="background:linear-gradient(90deg,darkgreen,darkgreen); -webkit-background-clip:text !important; -webkit-text-fill-color:transparent;">'''''What if the frictional force magnitude wasn't greater than the applied force?'''''</span><div class="mw-collapsible-content">If we were to have assumed that one of the other choices were correct, then the physical situation involving the police car would have been different.

* If we said that "The police car's applied force (from engine thrust)" had a greater magnitude, then the police car would of accelerated in the rightward direction in the video, continuing to speed up towards the right.

* If we said that "Neither" had a greater magnitude, then the friction and applied forces would of cancelled each other out. Although it will be later discussed as part of Newton's First Law in Topic 2.4, if we only have forces that cancel each other out, then the object won't accelerate in any direction. As such, choosing this would of implied that the police car's speed was unchanging.</div></div></div></div>
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====<big>Example 4: Propeller Planes</big> ====
</div>[[File:PlaneAppliedForce.webm|thumb|333x333px|A plane entering a gradual descent with a constant net force being applied on it.]]To expand the discussion of Newton's Second Law to quantitative situations, let's consider the map ''Maneuver Planes'' by ''TiM MELL0''. In this, players have the ability to influence a small airplane's motion by applying a force onto its surface.

Suppose that the player '''''Sam''''' (red color) is flying an airplane, initially applying a downwards force while moving horizontally towards the right. When ''Sam'' enters the dotted line track, he '''stops''' applying a force onto the airplane. Soon afterwards, the direction of the plane's motion approaches a steady, gradual descent. 

'''Assume that:'''

* The combined mass of the plane and player is <math>6{,}000. \; \text{kg}</math>
* The forces acting on the plane are unchanging while it is on the dotted line track.
* The <math>\vec F_{\text{grav}}</math> and the <math>\vec F_{\text{lift}}</math> forces occur in opposite directions and have an equal magnitude.
* The direction of <math>\vec F_{\text{engine}}</math> (force from plane engines) faces <math>5^{\circ} </math> below the +X Axis.
* The force magnitudes are <math>\vec F_{\text{lift}} = 10. \; \text{kN}</math>, <math>\vec F_{\text{grav}} = -10. \; \text{kN}</math>, and <math>\vec F_{\text{engine}} = 15. \; \text{kN}</math>.
(''Unit conversion:'' <math>1 \; \text{kN}</math> = <math>1{,}000 \; \text{N} </math>)
* The forces can be pictured as a free body diagram as provided below.

[[File:ForcesOnPlane.png|left|thumb|362x362px|A diagram showing the forces that are acting on a plane while in descent.]]
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'''Task:''' Calculate the the acceleration of the airplane in the horizontal direction while it is traveling on the dotted line track.
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<span style="background:black; -webkit-background-clip:text !important; -webkit-text-fill-color:transparent;">'''''Answer:'''''</span><div class="mw-collapsible-content">
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<math>
\vec a_{\text{net},\textbf{x}}= 2.5 \;\frac{\text{m}}{
\text{s}^2}</math> [2 sigfigs] <br> Note: The horizontal acceleration is being considered, not the overall acceleration.
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In order to solve this problem, we need to consider Newton's Second Law. However, we haven't been explicitly given <math>\vec F_{\text{net}} </math>. Instead, we have to use the listed information above to figure out <math>\vec F_{\text{net}} </math>.

We can find that <math>\vec F_{\text{lift}} </math> and <math>\vec F_{\text{grav}} </math> directly oppose and cancel each other out. As such, we can ignore these forces while finding <math>\vec F_{\text{net}} </math>. (They're also acting only in the vertical direction, making them not have an impact on the horizontal acceleration. However, we'll get to that shortly.)

As such, only <math>\vec F_{\text{engine}} </math> remains uncancelled. Therefore, <math>\vec F_{\text{net}} = \vec F_{\text{engine}} </math>. With this information, we can derive the overall acceleration of the plane, as shown below:
<math>\vec F_{\text{net}} = m\, \vec a_{\text{net}} </math> [Definition of Newton's Second Law]

<math>\vec a_{\text{net}} = \frac{\vec F_{\text{net}}}{m} </math> [Solving for overall acceleration vector.]

<math>\vec a_{\text{net}} = \frac{(15.)\cdot 10^3 \;\text{N}}{6000. \; \text{kg}} </math> [Substitution of known variables; <math> 15. \; \text{kN} \implies (15.)\cdot 10^3 \; \text{N} </math>].

<math>\vec a_{\text{net}} = 2.5 \; \frac{\text{N}}{\text{kg}} </math> [Division; <math> \frac{\text{N}}{\text{kg}} \implies \frac{\text{m}}{\text{s}^2}  </math>; 2 sigfigs].[[File:PlaneAccelerationVectorComponents.png|thumb|An overlay of the overall acceleration vector overall acceleration, <math>\vec a_{\text{net}}</math>, and its components on a still image of Sam's flight.|328x328px]]
However, '''we're not done just yet'''. This is because the task specifically asked for us to consider the <u>horizontal acceleration</u> of the plane, not the overall acceleration. As such, we need to use trigonometry to get the final answer. We can graph that the overall acceleration, <math>\vec a_{\text{net}}</math>, has a magnitude of <math>2.5 \; \frac{\text{m}}{\text{s}^2}</math> and has a direction angle that is <math>5^{\circ} </math> below the +X Axis. Concerning the direction angle of <math>\vec a_{\text{net}}</math>, we know this because of <math>\vec F_{\text{net}}</math>'s specified direction angle. Also, objects accelerate in the direction of its net force.


Using the acceleration vector <math>\vec a_{\text{net}}</math> and its components, we can construct a right angle triangle which a known angle <math>\theta</math> which equals <math>5^\circ</math>. Because <math>\vec a_{\text{net,x}}</math> is adjacent side to <math>\theta</math> and <math>\vec a_{\text{net}}</math> is the hypotenuse of the triangle, we can solve directly for <math>\vec a_{\text{net,x}}</math> as shown below:

<math>cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse side}}</math>[Trigonometric definition of the <math>cos(\theta)</math> function.]

<math>\text{adjacent side} = cos(\theta)(\text{hypotenuse side})</math> [Solving for the adjacent side length.]

<math>\vec a_{\text{net,x}}=cos(5^\circ)(2.5) \;\frac{\text{m}}{\text{s}^2}</math> [Substitution of variables.]

<math>\vec a_{\text{net,x}}= 2.49049 \ldots \;\frac{\text{m}}{
\text{s}^2}</math> [Computation (using degree mode).]

<math>\vec a_{\text{net,x}}= 2.5 \;\frac{\text{m}}{
\text{s}^2}</math> [2 sigfigs]

The plane is horizontally accelerating at a rate of <math>2.5 \;\frac{\text{m}}{
\text{s}^2}</math>.

To note, <math>\vec a_{\text{net,x}}</math> with 2 significant figures is equal to <math>\vec a_{\text{net}}</math>. This is because <math>\theta</math> is a small angle where <math>cos(\theta) = 0.99619\ldots \approx 1</math>.
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==== <big>Question 1: Frog Shrine</big> ====
</div>[[File:FrogShrineFinal.webm|thumb|419x419px|A player falls into a frog's tongue and is quickly pulled into its mouth by a strong force.]]Consider the map ''Frog shrine'' by ''O_dot''. In this map, players are engaging in direct combat with one another. One of the players accidentally touches a trap sensor (magenta colored), resulting in a decorative frog to suddenly launch its tongue. Shortly afterwards, '''''Spy Coder X''''' (blue colored) is snatched and pulled away to their demise.
'''Assume that:'''
* The frog's tongue is initially launched from its mouth with a velocity of <math>-2.0 \; \frac{\text{m}}{\text{s}}</math>. 
* The net force on the frog tongue, <math>\vec F_{\text{net}}</math>, is constant and has a magnitude of <math>500. \; \text{N}</math>
* The frog tongue's has a mass of <math>200. \; \text{kg}</math>
* ''Spy Coder X'' lands <math>0.20 \; \text{m}</math> away from the tip of the frog's tongue. This occurs <math>0.10 \; \text{s}</math> after the frog's tongue becomes fully extended.
* ''Spy Coder X'' is fully attached onto the frog's tongue upon contact.
'''Use the information provided above to answer the following:'''
<br>'''Part (a):'''<blockquote> Determine the direction in which the net force on the frog's tongue, <math>\vec F_{\text{net}}</math>, is acting towards. 
</blockquote>
'''Part (b):'''<blockquote>Using Newton's Second Law, calculate the acceleration of frog's tongue.</blockquote>
[[File:FBDFrogShrine.png|thumb|An unlabeled free body diagram of the forces that may be acting on the ''Spy Coder X'' specified under the player.
<br>
''Definitions of each of the mentioned forces:''
<br>
* Let <math>\vec F_{\text{grav}}</math> be the gravitational force that is acting on the player.
<br>
* Let <math>\vec F_{\text{contact}}</math> be the contact force from the frog's tongue that is acting on the player.
<br>
* Let <math>\vec F_{\text{tongue}}</math> be the horizontal force from the frog's tongue that is acting on the player. This is the net force that is acting on the frog's tongue.

|404x404px]]
'''Part (c):'''<blockquote>Draw a free body diagram of all of the forces acting on ''Spy Coder X'' after landing on the frog's tongue (as imaged on the right). Note that some of the forces mentioned in the image may not be needed.</blockquote>
'''Part (d):'''<blockquote>'''(i)''' Let the moment in which the frog's tongue be initially released be <math>t= 0.0 \; \text{s}</math>. At what time <math>t</math> does the frog's tongue return to its mouth? 
'''(ii)''' At what time <math>t</math> is the frog's tongue fully extended?

'''(iii)''' Calculate how much time elapses between ''Spy Coder X'' sticking to the frog's tongue and them entering the frog's mouth.
</blockquote>
''Consider discussing your solutions on this article's '''[[Talk:Physics Explained Through a Video Game/Newton's Second Law|Talk Page]]''', where you find help from others.''
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===References===
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