# Calculating Work From A Physics Viewpoint

These are my notes on work and the definite integral in physics.

Work is defined as the force, in the direction of a displacement, times that

displacement.

According to the definition, work is \(W=F_{o}x_{o}\). This is seen as a shaded

rectangular area. In calculus terminology, the work is the area under the curve.

The area under this curve is the area of the rectangle. The unit of work is the

joule (J=N*m).**Example 1**

Calculate the work performed by a constant force of 60 N acting in the direction

of a displacement of 3.0 m.

Solution:

The force is acting in the direction of the displacement, so the work is the

product of force times displacement.

\[W=F*x=60N*3.0m=180J\]

The force is in the direction of the displacement, and the work is simply the

product of force and displacement. The simplest way to obtain the component of

the force in the direction of the displacement is to form the dot product of

force and displacement. Using s as a general symbol for displacement, work is

defined as the dot product of the force vector times the displacement vector.

\[W = F * s\]

Work is a scalar, the dot product of force and displacement.**Example 2**

Calculate the work performed by a constant force of 40 N pushing a block a

distance of 3.0 m along a horizontal surface at an angle of 20 degrees.

Solution:

The work performed is calculated by:

\[W=F*s=40N \cos 20 * 3.0m=113J\]**Example 3**

Calculate the work done by the force of gravity acting on a 70 kg student

sliding down a 30 degree incline slide with a slant distance of 10 m.

Solution:

The first thing to calculate is the force due to gravity mg, which is 70 kg*9.8

m/s^2 = 686 N. This is the force of the student who is sliding down a 30 degree

incline that is 10 m long.

Calculate the force in the direction of the incline:

F=686 N cos 60 = 343 N

The work performed by the gravitational force is:

W = F * s = 343 N * 10 m = 3430 J

Now consider variable forces and look first at a force that is proportional to

x, the displacement, or F=kx, where k is a constant. This form of force is

encountered in springs where one unit of force compresses (or elongates) the

spring a certain distance and the next unit of force doubles the compression,

and so on.

The area under any curve of force versus distance can be approximated by taking

a sum of forces times increments of distances over the curve. In this instance,

take each increment as \(\Delta x\) and the associated forces as F1, F2, F3, up

to N forces and increments. In mathematical language, this sum of the forces

would be written as \(\sum F \Delta x\) where F represents the N discrete forces

measured at intervals of \(\Delta x\) from 0 to \(x_{0}\). This sum is a

collection of rectangles of height F1, F2, and so on until Fn and with width

\(\Delta x\).

Better approximations are obtained by increasing the number of intervals

(decreasing the width of each one). Geometrically, this limit of the sum

represents the area under the curve between x=0 and x=xo and physically

represents the work performed by this linear force over x. For a linear spring,

the work performed is the shaded triangle which is (1/2 base * width) or

\(kx^2/2\).**Example 4**

Calculate the work performed in compressing a spring with a force constant of

200 N/m the first 2.0 cm.

Solution:

The work performed is calculated by:

\[W=\frac{kx^2}{2}=\frac{200N}{m}*\frac{(30*10^{-2}m)^2}{2}=9.0*10^{-2}J\]**Example 5**

Calculate the work performed in compressing the spring the next 3.0 cm

Solution:

The general expression for the work performed in is \(kx^2/2\). The work to

compress the spring from 3.0 cm to 6.0 cm is the work to compress it 6.0 cm less

the work necessary to compress is 3.0 cm.

\[W=\frac{200N}{m}*\frac{1}{2}[(6.0*10^{-2}m)^2-(3.0*10^{-2}m)^2]=0.27J\]

This process of taking an infinite number of sums of infinitesimally small

increments defines an integral.

For the linear or Hooke's Law spring where F=kx, the integral becomes \(kx^2/2\)

which corresponds to the area under the curve of kx in the interval from 0 to

x0.

The area is the area of the triangle and is equal to \(kx^2/2\). We will not

develop the general technique for obtaining integrals of power law functions. We

will present a justification for the general formula for power law integrals.

Constant F \(F_{0}X_{0}\)

Linear kx \(kx^2/2\)

Quadratic kx \(kx^3/3\)

Power kx^n \(kx^{n+1}/n+1\)

The first two entries in the table are based on the area of a rectangle and a

triangle. The third entry, for the quadratic, follows the pattern of the

previous two and can be verified by numerically integrating the quadratic by

taking small increments and calculating the associated force and multiplying to

find the area of each little approximating rectangle. This process can also be

done for higher power curves to come to the conclusion shown in the last entry

in the table, the one for the general power law curve.

We now come to another definition of work as the integral of the dot product of

force and differential distance:

\[W=\int F*ds\]

Most of the situations encountered in an elementary course are for constant or

linear forces. The linear force is encountered in springs. Linear springs are

characterized by the constant k. The k in the equation F=kx has the units of

N/m. The work performed in compressing a kx spring is \(kx^2/2\). **Example 6**

What is the work performed in compressing a quadratic spring \(F=kx^2\) with

k=2000 N/m^2 from 0.30 m to 0.40 m?

Solution:

The force is in the direction of the displacement.

\[W = \int F*ds \int_{0.3}^{0.4} kx^2 dx = kx^3/3 = 2000N/m^2

1/3(0.4)^3-(0.3m)^3 = 25N*m = 25J\]

Notice the development of the definition of work. The first definition was the

product of the force in the direction of displacement times displacement. In

order to handle forces not in the direction of the displacement, the dot product

was introduced. Finally, for nonconstant forces, the integral was introduced,

giving the last definition.