Work Done By A Nonconstant Force: Difference between revisions

Revision as of 22:37, 9 April 2017

Claimed by Yunqing Jia (Spring 2017)

This page explains the significance and fundamental calculations of work done by non-constant forces. In addition, it provides multiple worked examples and analytical models will help readers develop a more thorough understanding.

The Main Idea

[1] [[2]] Work done by constant force of magnitude 'F' on a point that moves a displacement 'd' in the directional of the force is the product: W=F*d. The SI unit for work is the joule (J) which is equivalent to newton meter. Force is a relationship of the product of mass*acceleration of the object, while displace is a result of the change in final and starting position. It is important to note that no matter how large or small the magnitude of the force is, no work is done if there is no displacement. The formula W=F*d only holds true when a constant force is applied to the system.

When will we have to deal with non-constant forces? Why can't we just use the average force to make predictions?

Well, two common examples of non-constant forces are gravity and springs. Both forces depend on the position of the object and change constantly while the object is moving. For example, if we tried to calculate the Earth's orbital path around the sun using only the force and momentum from a single data point, the Earth would be moving away from the sun rather than circling around it, and humans would not have made it this far. If we tried to predict the behavior of a spring-mass system with a constant force, the spring would be either compressing or stretching forever instead of osillating between the two states. Therefore, we need a method to calculate the work done by non-constant forces in order to make preditions about the system.

Mathematical Model

Iterative calculations are used in order to calculate non-constant forces and predict an object's motion. Given initial and final states of a system under non-constant force, small displacement intervals should be used to calculate the object's trajectory. As mentioned in previous sections, the total amount of work done on a system equals the sum of works done by all individual forces, therefore, the total amount of work done can be calculated by the summation of the force on an object multiplied by the change in position in small increments.

[math]\displaystyle{ {{W}_{total} = {W}_{1} + {W}_{2} + {W}_{3} + ... + {W}_{n} = \overrightarrow{F}_{1}\bullet\overrightarrow{dr}_{1} + \overrightarrow{F}_{2}\bullet\overrightarrow{dr}_{2} + \overrightarrow{F}_{3}\bullet\overrightarrow{dr}_{3} + ... + \overrightarrow{F}_{n}\bullet\overrightarrow{dr}_{n}} }[/math]

[math]\displaystyle{ {{W}_{total} = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r}} }[/math]

However, this process is very repetitive and the calculation gets unnecessarily tedious. If we make the displacement intervals infinitesimally small, we are essentially taking the integral of force with respect to displacement, or finding the area under the curve of force by displacement.

[math]\displaystyle{ {{W}_{total} = \int\limits_{i}^{f}\overrightarrow{F}\bullet\overrightarrow{dr}} }[/math]

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.

Modeling Non-Constant Forces in VPython

example: a spring-mass system moving in a plane

[[<iframe src="https://trinket.io/embed/glowscript/c26c4c2637" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>]]

#intialize conditions
#calculation loop
  #calculate/update force at every time step
    L = ball.pos - spring.pos
    Lhat = norm(L)
    s = mag(L) - L0
    Fspring = -(ks * s) * Lhat
  #apply momentum principle
    ball.p = ball.p + (Fspring + Fgravity) * deltat
 #update positions
 #update time

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 = 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.

References

[3] [4]

Created by Justin Vuong

Edited by Chris Mickas

Edited by Yunqing Jia