Work - Revision history

Astovall8 at 05:14, 26 November 2018

2018-11-26T05:14:25Z

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	Written by: Jae Hyun Kim		Written by: Jae Hyun Kim

	~~'''Claimed~~ by ~~Deanna Berg Spring~~ 2018~~'''~~		(Contributed to by Alton Stovall- Fall 2018)

	==Work: The Main Idea==		==Work: The Main Idea==

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	==History==		==History==

	The ~~concept~~ of work ~~developed with the concept of conservation of energy~~. ~~In 1918~~, the ~~law~~ of ~~conservation energy was proved by~~ a ~~number~~ of ~~scientists~~, ~~and~~ the ~~concept~~ of ~~work was not developed by a specific person~~ at a ~~specific time~~.		The origins of energy and work are closely tied to observations made for heat or "phlogiston" as it was referred to centuries ago. Over time, the role of mathematics became more appreciated as a contributor to the underlying understanding of physical principles. Likewise, with the adoption of math at the forefront, a relationship between exerted force and displacement was discovered.

	== See also ==		== See also ==

Dberg6 at 02:31, 19 April 2018

2018-04-19T02:31:13Z

← Older revision		Revision as of 22:31, 18 April 2018
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	In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.		In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.
	<math>W = \vec(F) \bullet \Delta\vec(r) = F_x\~~Deltax~~ + F_y\~~Deltay~~ + F_z\~~Deltaz~~</math>		<math>W = \vec(F) \bullet \Delta\vec(r) = F_x\Delta x + F_y\Delta y + F_z\Delta z</math>

	An important thing to note is the sign of work:		An important thing to note is the sign of work:

Dberg6 at 02:30, 19 April 2018

2018-04-19T02:30:44Z

← Older revision		Revision as of 22:30, 18 April 2018
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	In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.		In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.
	W = \vec(F) \bullet \Delta\vec(r) = F_x\Deltax + F_y\Deltay + F_z\Deltaz</math>		<math>W = \vec(F) \bullet \Delta\vec(r) = F_x\Deltax + F_y\Deltay + F_z\Deltaz</math>

	An important thing to note is the sign of work:		An important thing to note is the sign of work:

Dberg6 at 02:30, 19 April 2018

2018-04-19T02:30:04Z

← Older revision		Revision as of 22:30, 18 April 2018
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	In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.		In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.
	W = \vec(F) \bullet \Delta\vec(r) = F_x\Deltax + F_y\Deltay + F_z\Deltaz		W = \vec(F) \bullet \Delta\vec(r) = F_x\Deltax + F_y\Deltay + F_z\Deltaz</math>

	An important thing to note is the sign of work:		An important thing to note is the sign of work:

Dberg6 at 02:29, 19 April 2018

2018-04-19T02:29:31Z

← Older revision		Revision as of 22:29, 18 April 2018
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	<math>W_g = -\Delta PE.</math>		<math>W_g = -\Delta PE.</math>

			In equations where vectors are present, a dot product method of multiplication is necessary. This is accomplished by multiplying x components, y components, and z components, then adding each x, y, and z component together.
			W = \vec(F) \bullet \Delta\vec(r) = F_x\Deltax + F_y\Deltay + F_z\Deltaz

			An important thing to note is the sign of work:
			#If the force is in the same direction as motion, work is positive
			#If the force is opposite the direction of motion, work is negative
			#If the force is perpendicular to the direction of motion, work is zero
			#If the displacement is zero, work is zero

	===A Computational Model===		===A Computational Model===

	The work done on the system by the ~~surrounding~~ is dependent on the system that on decides to choose. For example, if one decides to include ~~earth~~ as a part of the system while calculating the different forms of energies, the work done is 0. However, if one decides to count ~~earth~~ as the ~~surrounding~~, there is work done by ~~earth~~ but the potential energy from the previous case is eliminated.		The work done on the system by the surroundings is dependent on the system that one decides to choose. For example, if one decides to include Earth as a part of the system while calculating the different forms of energies, the work done is 0. However, if one decides to count Earth as the surroundings, there is work done by Earth but the potential energy from the previous case is eliminated.

	Because it is more difficult to use the momentum principle to update position of an object in motion, we typically ~~did~~ not use the ~~earth~~ as a surrounding. Rather, we ~~used~~ earth as part of the system and ~~calculated~~ gravitational force as net force. The example of the code is below;		Because it is more difficult to use the momentum principle to update position of an object in motion, we typically do not use the Earth as a surrounding. Rather, we use earth as part of the system and calculate gravitational force as net force. The example of the code is below;

	[[File:Code1J.png]]		[[File:Code1J.png]]
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	===Simple===		===Simple===

	'''A person picks up a 2kg box from the ground and lifts it up, moves 6 meters forward, and puts the box back down to the ground. What is the work done?'''		'''A person picks up a 2kg box from the ground and lifts it up, moves 6 meters forward, and puts the box back down on the ground. What is the work done?'''

	''Solution:''		''Solution:''
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	===Middling===		===Middling===

	'''Jack and Jill are maneuvering a 3200 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is 1.1 m/s. As the boat moves to position < 4, 0, 2 > m, Jack exerts a force of < -430, 0, 215 > N, and Jill exerts a force of < 150, 0, 300 > N.		'''Jack and Jill are maneuvering a 3200 kg boat near a dock. Initially the boat's position is \lt 2, 0, 3 \gt m and its speed is 1.1 m/s. As the boat moves to position \lt 4, 0, 2 \gt m, Jack exerts a force of \lt -430, 0, 215 \gt N, and Jill exerts a force of \lt 150, 0, 300 \gt N.

	a. How much work does Jack do?		a. How much work does Jack do?
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	a) The work done is addition of all the components (x,y and z) So the calculation goes:		a) The work done is addition of all the components (x,y and z) So the calculation goes:

	~~<math>Wjack~~ = -430 * (4-2) + 0 * 0 + 215 * (2-3) = -1075 ~~</math>~~		\ltmath\gtWjack = -430 * (4-2) + 0 * 0 + 215 * (2-3) = -1075

	b) The work done by Jill is calculated the same		b) The work done by Jill is calculated the same
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	A system of two identical blocks and spring are initially at rest. The blocks have mass M (labeled 1 and 2) are connected by an unstretched spring and slide on a frictionless surface. As shown in the upper diagram, a constant force F1 is applied to block 1 and a constant force F2 is applied to block 2. Initially the first block is located at X1 = 0. At a later time, the blocks are in a new position as indicated in the lower diagram. Block 1 has moved to the left and block 2 has moved to the right. At this final time, the system is moving to the left, vibrating, and the spring is stretched. In this diagram, the positive x-direction is to the right.		A system of two identical blocks and spring are initially at rest. The blocks have mass M (labeled 1 and 2) are connected by an unstretched spring and slide on a frictionless surface. As shown in the upper diagram, a constant force F1 is applied to block 1 and a constant force F2 is applied to block 2. Initially the first block is located at X1 = 0. At a later time, the blocks are in a new position as indicated in the lower diagram. Block 1 has moved to the left and block 2 has moved to the right. At this final time, the system is moving to the left, vibrating, and the spring is stretched. In this diagram, the positive x-direction is to the right.



	[[File:hardworkproblem.jpg]]		[[File:hardworkproblem.jpg]]

Dberg6 at 20:28, 18 April 2018

2018-04-18T20:28:57Z

← Older revision		Revision as of 16:28, 18 April 2018
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	From this we can conclude that:		From this we can conclude that:

	<math>W = -\Delta PE.</math>		<math>W_g = -\Delta PE.</math>

	===A Computational Model===		===A Computational Model===

Dberg6 at 20:28, 18 April 2018

2018-04-18T20:28:18Z

← Older revision		Revision as of 16:28, 18 April 2018
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	[[File:workdiagram.jpg]]		[[File:workdiagram.jpg]]

	~~It is important to note that when~~ the direction of the force is equal to the direction of the displacement, the "theta" aspect of this equation is equal to 0, meaning that cos(0) will just be 1 ~~and~~ the formula ~~can further be simplified to W = Fd~~.		In reality, if the direction of the force is equal to the direction of the displacement, the "theta" aspect of this equation is equal to 0, meaning that cos(0) will just be 1. So, technically in certain situations this is the same as our generic formula.

	2) When the work is done by a non-constant force:		2) When the work is done by a non-constant force:

	<math> W=\int\limits_{i}^{f}\overrightarrow{F}\bullet\overrightarrow{dr} = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} </math>		<math> W=\int\limits_{i}^{f}\overrightarrow{F}\bullet\Delta\overrightarrow{r} = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} </math>

	This might be a little easier to understand for those who have exposure to calculus. The work done by a non-constant force is the summation of the products of the forces acting on a system and the change in position of the system. This can be denoted by an integral; work is equal to the integral of the force with respect to change in displacement.		This might be a little easier to understand for those who have exposure to calculus. The work done by a non-constant force is the summation of the products of the forces acting on a system and the change in position of the system. This can be denoted by an integral; work is equal to the integral of the force with respect to change in displacement. In a computational, or coding explanation, this is denoted in loops.

	3) When work is done by gravity:		3) When work is done by gravity:
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	<math>W = F_g (y_2 - y_1) = F_g\Delta y = - mg\Delta y</math>		<math>W = F_g (y_2 - y_1) = F_g\Delta y = - mg\Delta y</math>

	Notice that this is exact same formula as ~~explained in number one~~, when the work is done by a constant force, except it is more specified~~. Theta was neglected because the direction of the gravitational pull was the same as change in height~~. Notice that this is also similar to the formula for change in potential energy, which is:		Notice that this is exact same formula as the generic formula, when the work is done by a constant force, except it is more specified. Notice that this is also similar to the formula for change in potential energy, which is:

	<math>\Delta PE = mg\Delta y</math>.		<math>\Delta PE = mg\Delta y</math>.

Dberg6 at 20:22, 18 April 2018

2018-04-18T20:22:40Z

← Older revision		Revision as of 16:22, 18 April 2018
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	1) When the work is done by a constant force at an angle to the displacement:		1) When the work is done by a constant force at an angle to the displacement:

	<math> W = \overrightarrow{F} \bullet \~~delta~~\overrightarrow{r} \cos(\theta)</math>		<math> W = \overrightarrow{F} \bullet \Delta\overrightarrow{r} \cos(\theta)</math>

	[[File:workdiagram.jpg]]		[[File:workdiagram.jpg]]

Dberg6 at 20:22, 18 April 2018

2018-04-18T20:22:10Z

← Older revision		Revision as of 16:22, 18 April 2018
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	1) When the work is done by a constant force at an angle to the displacement:		1) When the work is done by a constant force at an angle to the displacement:

	<math> W = \overrightarrow{F} \bullet \overrightarrow{dr} \cos(\theta).</math>		<math> W = \overrightarrow{F} \bullet \delta\overrightarrow{r} \cos(\theta)</math>

	[[File:workdiagram.jpg]]		[[File:workdiagram.jpg]]

Dberg6 at 20:21, 18 April 2018

2018-04-18T20:21:26Z

← Older revision		Revision as of 16:21, 18 April 2018
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	In physics, it is easiest to think of work as energy produced by a force acting through a specific distance. From this definition, it can be seen that work is used in relation to the Energy Principle; therefore, the units of work are Joules (J). Additionally, from this definition, force (F) and distance (d) are the two major contributors to determining work. However, using the word "distance" can be misleading; the actual formula uses the displacement (∆r).		In physics, it is easiest to think of work as energy produced by a force acting through a specific distance. From this definition, it can be seen that work is used in relation to the Energy Principle; therefore, the units of work are Joules (J). Additionally, from this definition, force (F) and distance (d) are the two major contributors to determining work. However, using the word "distance" can be misleading; the actual formula uses the displacement (∆r).

	Some important things to note ~~are the~~ directions of motion and force in relation to each other. This concept will be looked into more in depth in the Mathematical Model section.		Some important things to note:
			#The directions of motion and force in relation to each other. This concept will be looked into more in depth in the Mathematical Model section.
			#With no displacement, no work is done.





	In physics, work refers to the "efficiency" of the force. A force is doing "work" when acting upon a matter or a particle if it causes the matter to move or change in displacement. Therefore, ''work is the product of force and distance. '' Overall, the formula appears very simple, but there is an underlying reasoning as to why it is what it is.

	It is important to consider the three so called "ingredients" when determining work. Those ingredients are the force, displacement and a cause. The force must "cause" the change in displacement in order for it to actually be considered as work. With no displacement, no work is done.

	[[File:workdone.jpg]]		[[File:workdone.jpg]]

	While work may seem like a simple concept, it is nonetheless extremely essential in physics calculations as it is a core aspect of a fundamental principle, the conservation of energy. ~~This will be further discussed in the "connections" part where the individual source of energy work takes part to take~~ role in the transfer of energy from the ~~surrounding~~ to a system ~~or from~~ the ~~system to the surrounding to contribute to the conservation of energy~~.		While work may seem like a simple concept, it is nonetheless extremely essential in physics calculations as it is a core aspect of a fundamental principle, the conservation of energy. Work plays a major role in the transfer of energy from the surroundings to a system.This will be further discussed in the "connections" part.

	===A Mathematical Model===		===A Mathematical Model===
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	<math>W = \vec{F} \bullet ∆\vec{r}</math>		<math>W = \vec{F} \bullet ∆\vec{r}</math>

			Depending on the conditions of a specific problem, this formula can be modified to better fit what is being asked.


			1) When the work is done by a constant force at an angle to the displacement:

	~~There are a few mathematical formulas to determine or calculate the amount of work done by a force. These formulas are a bit different from situation to situation.~~		<math> W = \overrightarrow{F} \bullet \overrightarrow{dr} \cos(\theta).</math>

	~~First of all, the formula for work is different according to the consistency of the force.~~

	~~1) When the work is done by a constant force:~~

	<math> W=\overrightarrow{F}\bullet\overrightarrow{dr}\cos\theta.</math>

	[[File:workdiagram.jpg]]		[[File:workdiagram.jpg]]