<< Chapter < Page Chapter >> Page >

Check Your Understanding Can Earth’s gravity ever be a constant force for all paths?

No, it’s only approximately constant near Earth’s surface.

Got questions? Get instant answers now!

Work done by forces that vary

In general, forces may vary in magnitude and direction at points in space, and paths between two points may be curved. The infinitesimal work done by a variable force can be expressed in terms of the components of the force and the displacement along the path,

d W = F x d x + F y d y + F z d z .

Here, the components of the force are functions of position along the path, and the displacements depend on the equations of the path. (Although we chose to illustrate dW in Cartesian coordinates, other coordinates are better suited to some situations.) [link] defines the total work as a line integral, or the limit of a sum of infinitesimal amounts of work. The physical concept of work is straightforward: you calculate the work for tiny displacements and add them up. Sometimes the mathematics can seem complicated, but the following example demonstrates how cleanly they can operate.

Work done by a variable force over a curved path

An object moves along a parabolic path y = ( 0.5 m −1 ) x 2 from the origin A = ( 0 , 0 ) to the point B = ( 2 m, 2 m ) under the action of a force F = ( 5 N/m ) y i ^ + ( 10 N/m ) x j ^ ( [link] ). Calculate the work done.

A graph of y in meters versus x in meters is shown. A parabolic path labeled as y of x starts at 0, 0 and curves up and to the right. The point (2, 2) is on the parabola. Vector F of x, y is shown at a point between the origin and coordinate 2, 2. Vector F points to the right and up, at some angle to the curve y of x.
The parabolic path of a particle acted on by a given force.

Strategy

The components of the force are given functions of x and y . We can use the equation of the path to express y and dy in terms of x and dx ; namely,

y = ( 0.5 m −1 ) x 2 and d y = 2 ( 0.5 m −1 ) x d x .

Then, the integral for the work is just a definite integral of a function of x .

Solution

The infinitesimal element of work is

d W = F x d x + F y d y = ( 5 N/m ) y d x + ( 10 N/m ) x d y = ( 5 N/m ) ( 0.5 m 1 ) x 2 d x + ( 10 N/m ) 2 ( 0.5 m 1 ) x 2 d x = ( 12.5 N/m 2 ) x 2 d x .

The integral of x 2 is x 3 / 3 , so

W = 0 2 m ( 12.5 N/m 2 ) x 2 d x = ( 12.5 N/m 2 ) x 3 3 | 0 2 m = ( 12.5 N/m 2 ) ( 8 3 ) = 33.3 J .

Significance

This integral was not hard to do. You can follow the same steps, as in this example, to calculate line integrals representing work for more complicated forces and paths. In this example, everything was given in terms of x - and y -components, which are easiest to use in evaluating the work in this case. In other situations, magnitudes and angles might be easier.

Got questions? Get instant answers now!

Check Your Understanding Find the work done by the same force in [link] over a cubic path, y = ( 0.25 m −2 ) x 3 , between the same points A = ( 0 , 0 ) and B = ( 2 m, 2 m ) .

W = 35 J

Got questions? Get instant answers now!

You saw in [link] that to evaluate a line integral, you could reduce it to an integral over a single variable or parameter. Usually, there are several ways to do this, which may be more or less convenient, depending on the particular case. In [link] , we reduced the line integral to an integral over x , but we could equally well have chosen to reduce everything to a function of y . We didn’t do that because the functions in y involve the square root and fractional exponents, which may be less familiar, but for illustrative purposes, we do this now. Solving for x and dx , in terms of y , along the parabolic path, we get

x = y / ( 0.5 m −1 ) = ( 2 m ) y and d x = ( 2 m ) × 1 2 d y / y = d y / ( 2 m −1 ) y .

The components of the force, in terms of y , are

F x = ( 5 N/m ) y and F y = ( 10 N/m ) x = ( 10 N/m ) ( 2 m ) y ,

Questions & Answers

what is the guess theorem
Monu Reply
viva question and answer on practical youngs modulus by streching
Akash Reply
send me vvi que
rupesh
a car can cover a distance of 522km on 36 Liter's of petrol, how far can it travel on 14 liter of petrol.
Isaac
whats a two dimensional force
Jimoh Reply
what are two dimensional force?
Ahmad
Where is Fourier Theorem?
Atul Reply
what is Boyle's law
Amoo Reply
Boyle's law state that the volume of a given mass of gas is inversely proportion to its pressure provided that temperature remains constant
Abe
how do I turn off push notifications on this crap app?
Huntergirl
what is the meaning of in.
CHUKWUMA Reply
In means natural logarithm
Elom
is dea graph for cancer caliper experiment using glass block?
Bako
identity of vectors?
Choudhry Reply
what is defined as triple temperature
Prince Reply
Triple temperature is the temperature at which melting ice and boiling water are at equilibrium
njumo
a tire 0.5m in radius rotate at constant rate 200rev/min. find speed and acceleration of small lodged in tread of tire.
Tahira Reply
hmm
Ishaq
100
Noor
define the terms as used in gravitational mortion 1:earth' satellites and write two example 2:parking orbit 3:gravitation potential 4:gravitation potential energy 5:escping velocity 6:gravitation field and gravitation field strength
Malima Reply
can gravitational force cause heat?
SANT
yes
Kawshik
_2/3 ÷34
Isaac
what larminar flow
Rajab Reply
smooth or regular flow
Roha
Hii
Sadiq
scalar field define with example
Malik Reply
what is displacement
Isaac Reply
the change in the position of an object in a particular direction is called displacement
Noor
The physical quantity which have both magnitude and direction are known as vector.
Malik
good
Noor
Describe vector integral?
Malik
define line integral
Malik
Examples on how to solve terminal velocity
Louis Reply
what is Force?
Bibas Reply
ans:loading...
Lumai
the sideways pressure exerted by fluid is equal and canceled out.how and why?
Chaurasia
Practice Key Terms 2

Get the best University physics vol... course in your pocket!





Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask