Work

Author: Jae Hyun Kim

Claimed by Osman Ghandour March 21 2017

Work: The Main Idea

Outside of physics, people often use the word "work" to describe schoolwork, jobs, exercise, or anything that essentially requires effort to complete. In general terms, the word "work" is used extensively, but with good meaning.

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:Workisdone.png

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.

A Mathematical Model

There are a few mathematical formulas to determine or calculate the amount of work done by the force. These formulas could be a bit different from situations to situations.

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]\displaystyle{ W=\overrightarrow{F}\bullet\overrightarrow{dr}\cos\theta. }[/math]

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.

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

[math]\displaystyle{ W=\int\limits_{i}^{f}\overrightarrow{F}\bullet\overrightarrow{dr} = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} }[/math]

This might be a little easier to understand for those who have exposure to calculus. In more of simpler words, the work done by a non-constant force is the product of the summation of the forces that acts upon a system and the change in position of the system. This could be denoted by the usage of a calculus term integral; work is equal to the integral of the force in respect to change in displacement.

3) When work is done by gravity:

[math]\displaystyle{ 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:

[math]\displaystyle{ \Delta PE = mg\Delta y }[/math].

From this we can conclude that:

[math]\displaystyle{ W = -\Delta PE. }[/math]

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.

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;

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

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?

Solution:

The essence of this question is to apply the fact that the direction and the force has to be the same. Did the person apply a force in x direction? No. Thus, the x component of work is 0.

What about the Y component? Yes, the person did work against gravity. The magnitude of the work done when he pulled the box up can be calculated by the product of the mass, the gravitational acceleration, and the change in height (y). However, the overall change in displacement in terms of Y is 0, because he puts it back down.

Thus, the answer for this simple problem is 0.

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.

a. How much work does Jack do?

b. How much work does Jill do?

c. What is the final speed of the boat?

Solutions:

a) The work done is addition of all the components (x,y and z) So the calculation goes:

[math]\displaystyle{ Wjack = -430 * (4-2) + 0 * 0 + 215 * (2-3) = -1075 }[/math]

b) The work done by Jill is calculated the same

[math]\displaystyle{ Wjill = 150 * (4-2) + 0 * 0 + 300 * (2-3) = 0 }[/math]

c) To calculate the final speed, one must use the momentum principle. The two forms of energy present in this calculation is the kinetic energy and the work done. Thus,

[math]\displaystyle{ Ki + W = Kf }[/math] [math]\displaystyle{ KE = 1/2 * mass * V^2 }[/math]

Thus,

[math]\displaystyle{ Ki = 0.5 * 3200 * 1.1 ^ 2 = 1936J }[/math]

[math]\displaystyle{ Kf = 1936 - 1075 = 861J }[/math]

[math]\displaystyle{ Kf = 1/2 * mass * Vf^2 = 861 }[/math]

[math]\displaystyle{ Vf = 0.73 }[/math]

Difficult

a. How far did the center of mass of the extended system move

The main idea is that we have to calculate the change in displacement of the center of the mass not the force. Thus, we have to get the initial center of the mass, which is between x1 and x2, and the final position of the center of the mass, which is between x3 and x4. Getting the difference between two can yield the movement of the center of the mass.

b. Use the point particle system to determine the velocity of the center of mass for the system.

In the point particle system, we only take account of the kinetic transnational energy, and this is equivalent to the work done. The work is simply the change in displacement multiplied by the force. In this case, because there are two forces F1 and F2, one must calculate the edition of the both to get the net force.

C. Use the extended system to determine the final vibrational energy of the extended system.

In the extended system, we take acount of the kinetic transitional energy and the vibrational, of which sum is equivalent to the work.

Connectedness

1. How does this topic contribute to the overall physics?

The concept of work, seemingly an individual concept, is actually an essential part of the fundamental principle, the energy principle. According to the energy principle, the energy of a system is conserved unless the surrounding exerts a force upon the system. This force, when it causes the system to move, is called the work done by surrounding. Thus, it is important to consider work as part of the transitions in energy. Thus, when calculating the final and initial states of the potential and kinetic energy of a system after the work has been done, one can use the following formula:

2. How is this topic connected to something that you are interested in?

I a first year biomedical engineering major so I had to think a bit deeper to answer this question. I have concluded that the concept of work is actually prevalent throughout our body physiology. For example, in order for kinetic energy and potential energy of the muscles cells in our body to be changed, there must be an external force acting upon it changing its displacement. This concept is actually used especially in tissue engineering or biomechanics, which I would love to delve into in the future.

3. Is there an interesting industrial application?

Biomedical engineering is rather distant from the concept of work from other majors. Mechanical Engineering majors have a variety of interesting industrial applications that utilizes the concept of work. For example, when designing the engines of an airplane, mechanical engineers must take account of the work done by the engines of the airplane to win the gravitational pull that would attempt to crash the airplane. Without this concept of work, airplanes would not work.

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.

See also

The Energy Principle

Kinetic Energy

Thermal Energy

Conservation of Energy

Further reading

http://www.physicsclassroom.com/class/energy

External links

https://www.khanacademy.org/science/physics/work-and-energy

Khan academy does a great job in thoroughly explaining the concept of work using visualizations.

References

http://www.physicsclassroom.com/class/energy