Work Done By A Nonconstant Force: Difference between revisions
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===Example 2=== | ===Example 2=== | ||
We know that the formula for force is <math> F=ks </math>, where <math> s </math> is the distance the spring is stretched. If we integrate this with respect to <math> s </math>, we find that <math> W=.5ks^2 </math> is the formula for work. | |||
<math> W=\int\limits_{i}^{f}\overrightarrow{ | <math> W=\int\limits_{i}^{f}\overrightarrow{k}\bullet\overrightarrow{ds} = .5ks^2 </math> | ||
Say that we want to find the work done by a horizontal spring with spring constant k=100 N/m as it moves an object 15 cm. Using the formula W=.5ks2 that we derived from F=ks, we can calculate that the work done by the spring is 1.125 J. | |||
<math> W=\int\limits_{0}^{ | <math> W=\int\limits_{0}^{15}100\bullet\overrightarrow{ds}=.5ks^2=.5(100)(0.15^2)=1.125 J </math> | ||
===Example 3=== | ===Example 3=== | ||
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First, we must recall the formula for gravitational force. | First, we must recall the formula for gravitational force. | ||
Because G, M, and m are constants, we can remove them from the integral. We also know that the integral of 1/ | Because <math> G </math>, <math> M </math>, and <math> m </math> are constants, we can remove them from the integral. We also know that the integral of <math> -1\over r^2 </math> is <math> 1\over r </math>. We then must calculate the integral of <math> –GMm\over r^2 </math> from the initial radius of the asteroid, <math> R </math>, to the radius of the earth, <math> r </math>. | ||
<math> W=-GMm\bullet\int\limits_{R}^{r}{-1\over r^2}\bullet dr </math> | |||
<math> W=-GMm\bullet({1\over r}-{1\over R}) </math> | |||
Our answer will be positive because the force done by the earth on the asteroid and the direction of the asteroid's displacement are the same. | |||
==Connectedness== | ==Connectedness== | ||
Revision as of 11:11, 17 April 2016
Claimed by Chris Mickas
This page will help students understand how to calculate the work done by a non constant force.
The Main Idea
Basic calculations of work can be solved with a simple formula: force times displacement. However, this formula only works when work is constant. Calculus is essential for the calculation of work in many cases because force is not always a constant. In cases such as springs and gravity, for example, the force applied varies depending on location. By integrating, or finding the area under the curve of force by displacement, we can calculate work without having to use inaccurate approximations.
A Mathematical Model
[math]\displaystyle{ W=\int\limits_{i}^{f}\overrightarrow{F}\bullet\overrightarrow{dr} = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} }[/math]
This means that the work is equal to the integral of the function of the force with respect to the change in the objects position. This is also the same as the summation of the force on an object multiplied by the change in position.
A Computational Model
<https://trinket.io/glowscript/49f7c0f35f>
This model shows both the total work and the work done by a spring on a ball attached to a vertical spring. The work done by the spring oscillates because the work is negative when the ball is moving away from the resting state and is positive when the ball moves towards it.
Because gravity causes the ball’s minimum position to be further from the spring’s resting length than its maximum position could be, the work is more negative when the ball approaches its minimum height.
The code works by using small time steps of 0.01 seconds and finding the work done in each time step. Work is the summation of all of the work done in each time step, so another step makes sure the value for work is cumulative.
Examples
Example 1
A box is pushed to the East, 5 meters by a force of 40 N, then it is pushed to the north 7 meters by a force of 60 N. Calculate the work done on the box.
[math]\displaystyle{ W = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} }[/math]
[math]\displaystyle{ W = 40N \bullet\ 5m + 60N \bullet\ 7m }[/math]
[math]\displaystyle{ W = 40N \bullet\ 5m + 60N \bullet\ 7m }[/math]
[math]\displaystyle{ W = 620 J }[/math]
Example 2
We know that the formula for force is [math]\displaystyle{ F=ks }[/math], where [math]\displaystyle{ s }[/math] is the distance the spring is stretched. If we integrate this with respect to [math]\displaystyle{ s }[/math], we find that [math]\displaystyle{ W=.5ks^2 }[/math] is the formula for work.
[math]\displaystyle{ W=\int\limits_{i}^{f}\overrightarrow{k}\bullet\overrightarrow{ds} = .5ks^2 }[/math]
Say that we want to find the work done by a horizontal spring with spring constant k=100 N/m as it moves an object 15 cm. Using the formula W=.5ks2 that we derived from F=ks, we can calculate that the work done by the spring is 1.125 J.
[math]\displaystyle{ W=\int\limits_{0}^{15}100\bullet\overrightarrow{ds}=.5ks^2=.5(100)(0.15^2)=1.125 J }[/math]
Example 3
The earth does work on an asteroid approaching from an initial distance r. How much work is done on the asteroid by gravity before it hits the earth’s surface?
First, we must recall the formula for gravitational force.
Because [math]\displaystyle{ G }[/math], [math]\displaystyle{ M }[/math], and [math]\displaystyle{ m }[/math] are constants, we can remove them from the integral. We also know that the integral of [math]\displaystyle{ -1\over r^2 }[/math] is [math]\displaystyle{ 1\over r }[/math]. We then must calculate the integral of [math]\displaystyle{ –GMm\over r^2 }[/math] from the initial radius of the asteroid, [math]\displaystyle{ R }[/math], to the radius of the earth, [math]\displaystyle{ r }[/math].
[math]\displaystyle{ W=-GMm\bullet\int\limits_{R}^{r}{-1\over r^2}\bullet dr }[/math]
[math]\displaystyle{ W=-GMm\bullet({1\over r}-{1\over R}) }[/math]
Our answer will be positive because the force done by the earth on the asteroid and the direction of the asteroid's displacement are the same.
Connectedness
I am most interested in the types of physics problems that accurately model real world situations. Some forces, like gravity near the surface of the earth and some machine-applied forces, are constant. However, most forces in the real world are not.
Because of this, calculating work for non-constant forces is essential to mechanical engineering. For example, when calculating work done by an engine over a distance, the force applied by the engine can vary depending on factors such as user controls.
On an industrial level, the work needed to fill and empty tanks depends on the weight of the liquid, which varies as the tanks fill and empty. Energy conversion in hydroelectric dams depends on the work done by water against turbines, which depends on the flow of water. Windmills work in the same way.
History
Gaspard-Gustave de Coriolis, famous for discoveries such as the Coriolis effect, is credited with naming the term “work” to define force applied over a distance. Later physicists combined this concept with Newtonian calculus to find work for non-constant forces.
See also
References
Created by Justin Vuong