Work Done By A Nonconstant Force: Difference between revisions

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'''Claimed by Chris Mickas'''
= Work Done By A Nonconstant Force =
'''Claimed by Matt McCrory – Spring 2025'''


This page will help students understand how to calculate the work done by a non constant force.
This page explains how to calculate work done when the force applied is not constant. It includes conceptual explanations, worked examples, mathematical and computational models, and embedded simulations to make this concept easier to understand.


==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.
== The Main Idea ==
Before we understand nonconstant force, let's review constant force.


===A Mathematical Model===
For constant force:
: '''Work = Force × Distance'''
: <math>W = F \cdot d</math>


<math> W=\int\limits_{i}^{f}\overrightarrow{F}\bullet\overrightarrow{dr} = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} </math>
[[File:ConstantForce.png|thumb|center|300px|Work as the area under a constant force graph]]


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.
In real-life, however, forces often vary over distance. In that case, we use:
: <math>W = \int_{x_1}^{x_2} F(x) \, dx</math>


===A Computational Model===
This integral calculates the total work as the area under the curve on a Force vs. Distance graph.


<https://trinket.io/glowscript/49f7c0f35f>
== Mathematical Model ==
Work done by a varying force is found by breaking the motion into tiny intervals:


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.
: <math>W = \sum \vec{F}_i \cdot \Delta \vec{r}_i</math>


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.
As the interval becomes very small, it becomes a definite integral:
: <math>W = \int \vec{F} \cdot d\vec{r}</math>


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.
=== Spring Example ===
If <math>F = kx</math>, we derive:
: <math>W = \int_0^x kx \, dx = \frac{1}{2}kx^2</math>


[[File:WorkIntegral.png|thumb|center|300px|Work done by a spring force]]


==Examples==
== Computational Model ==
Computational models can approximate work using many tiny time steps. Below is Python code modeling a vertical spring in VPython:


===Example 1===
<syntaxhighlight lang="python">
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.
#initialize conditions
L = ball.pos - spring.pos
Lhat = norm(L)
s = mag(L) - L0
Fspring = -(ks * s) * Lhat


<math> W = \sum\overrightarrow{F}\bullet\Delta\overrightarrow{r} </math>
#momentum principle
ball.p = ball.p + (Fspring + Fgravity) * deltat
</syntaxhighlight>


<math> W = 40N \bullet\ 5m + 60N \bullet\ 7m </math>


<math> W = 40N \bullet\ 5m + 60N \bullet\ 7m </math>
== Interactive Model ==
Try out this Trinket simulation of spring motion: 
[https://trinket.io/glowscript/49f7c0f35f View the simulation on Trinket]


<math> W = 620 J </math>
== Examples ==
=== Simple ===
'''Question:''' 
A box is pushed 10 m east by a 40 N force, then 8 m north by a 60 N force. 
'''Solution:''' 
: <math>W = 40 \cdot 10 + 60 \cdot 8 = 880 \, J</math>


===Example 2===
=== Middling ===
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.
'''Question:''' 
A spring with <math>k = 70 \, N/m</math> is stretched 10 cm.
[[File:Middle1.JPG|thumb|center|200px|Spring stretching setup]]


<math> W=\int\limits_{i}^{f}\overrightarrow{k}\bullet\overrightarrow{ds} = .5ks^2 </math>
'''Solution:''' 
: <math>W = \frac{1}{2} k x^2 = \frac{1}{2}(70)(0.1)^2 = 0.35 \, J</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.
=== Difficult ===
'''Question:''' 
How much work is done by Earth’s gravity on an asteroid falling from distance <math>d</math> to radius <math>R</math>?


<math> W=\int\limits_{0}^{15}100\bullet\overrightarrow{ds}=.5ks^2=.5(100)(0.15^2)=1.125 J </math>
'''Solution:''' 
Start with Newton’s law of gravitation:
: <math>F = \frac{GMm}{r^2}</math>


===Example 3===
Then integrate:
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?
: <math>W = \int_R^d \frac{GMm}{r^2} \, dr = GMm \left( \frac{1}{R} - \frac{1}{d} \right)</math>


First, we must recall the formula for gravitational force.
== Connectedness ==
Understanding work by nonconstant forces is key in many fields:


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>.
* '''Springs''': Used in trampolines, shock absorbers, and mechanical pens 
* '''Engineering''': Fluid tanks fill unevenly, requiring nonconstant work 
* '''Energy''': Hydroelectric turbines rely on variable water flow 
* '''Space physics''': Rockets and satellites feel variable gravity


<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>
== History ==
Gaspard-Gustave de Coriolis was the first to define "work" as force over distance. Later physicists used calculus to model work by nonconstant forces.


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.
== Further Reading & External Links ==


==Connectedness==
=== Book ===
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.  
* Chabay & Sherwood – ''Matter and Interactions'' (4th ed.)


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.
=== Articles ===
* [https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/13%3A_Energy_Kinetic_Energy_and_Work/13.05%3A_Work_done_by_Non-Constant_Forces Nonconstant Force]
* [https://trinket.io/glowscript/49f7c0f35f Iterative Spring-Mass Simulation]


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.
=== Simulations ===
 
* [https://trinket.io/glowscript/49f7c0f35f Spring-Mass Trinket Model]
 
==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 ==
 
[[Work]]
 
==References==
 
[http://www.britannica.com/biography/Gustave-Gaspard-Coriolis]
[https://en.wikibooks.org/wiki/FHSST_Physics/Work_and_Energy/Work]
[https://trinket.io/glowscript/31d0f9ad9e]
 
[[Category:Energy]]
 
Created by Justin Vuong

Latest revision as of 22:24, 22 April 2025

Work Done By A Nonconstant Force

Claimed by Matt McCrory – Spring 2025

This page explains how to calculate work done when the force applied is not constant. It includes conceptual explanations, worked examples, mathematical and computational models, and embedded simulations to make this concept easier to understand.


The Main Idea

Before we understand nonconstant force, let's review constant force.

For constant force:

Work = Force × Distance
[math]\displaystyle{ W = F \cdot d }[/math]
Work as the area under a constant force graph

In real-life, however, forces often vary over distance. In that case, we use:

[math]\displaystyle{ W = \int_{x_1}^{x_2} F(x) \, dx }[/math]

This integral calculates the total work as the area under the curve on a Force vs. Distance graph.

Mathematical Model

Work done by a varying force is found by breaking the motion into tiny intervals:

[math]\displaystyle{ W = \sum \vec{F}_i \cdot \Delta \vec{r}_i }[/math]

As the interval becomes very small, it becomes a definite integral:

[math]\displaystyle{ W = \int \vec{F} \cdot d\vec{r} }[/math]

Spring Example

If [math]\displaystyle{ F = kx }[/math], we derive:

[math]\displaystyle{ W = \int_0^x kx \, dx = \frac{1}{2}kx^2 }[/math]
Work done by a spring force

Computational Model

Computational models can approximate work using many tiny time steps. Below is Python code modeling a vertical spring in VPython:

<syntaxhighlight lang="python">

  1. initialize conditions

L = ball.pos - spring.pos Lhat = norm(L) s = mag(L) - L0 Fspring = -(ks * s) * Lhat

  1. momentum principle

ball.p = ball.p + (Fspring + Fgravity) * deltat </syntaxhighlight>


Interactive Model

Try out this Trinket simulation of spring motion: View the simulation on Trinket

Examples

Simple

Question: A box is pushed 10 m east by a 40 N force, then 8 m north by a 60 N force. Solution:

[math]\displaystyle{ W = 40 \cdot 10 + 60 \cdot 8 = 880 \, J }[/math]

Middling

Question: A spring with [math]\displaystyle{ k = 70 \, N/m }[/math] is stretched 10 cm.

Spring stretching setup

Solution:

[math]\displaystyle{ W = \frac{1}{2} k x^2 = \frac{1}{2}(70)(0.1)^2 = 0.35 \, J }[/math]

Difficult

Question: How much work is done by Earth’s gravity on an asteroid falling from distance [math]\displaystyle{ d }[/math] to radius [math]\displaystyle{ R }[/math]?

Solution: Start with Newton’s law of gravitation:

[math]\displaystyle{ F = \frac{GMm}{r^2} }[/math]

Then integrate:

[math]\displaystyle{ W = \int_R^d \frac{GMm}{r^2} \, dr = GMm \left( \frac{1}{R} - \frac{1}{d} \right) }[/math]

Connectedness

Understanding work by nonconstant forces is key in many fields:

  • Springs: Used in trampolines, shock absorbers, and mechanical pens
  • Engineering: Fluid tanks fill unevenly, requiring nonconstant work
  • Energy: Hydroelectric turbines rely on variable water flow
  • Space physics: Rockets and satellites feel variable gravity


History

Gaspard-Gustave de Coriolis was the first to define "work" as force over distance. Later physicists used calculus to model work by nonconstant forces.

Further Reading & External Links

Book

  • Chabay & Sherwood – Matter and Interactions (4th ed.)

Articles

Simulations