Work

Work is said to be done by a force acting on a body, provided the body has displaced actually in any direction except in a direction perpendicular to the direction of the force.

For example, when a person carrying a briefcase moves on a horizontal road, he is not performing any work as distance moved in the direction perpendicular to the force applied. Similarly, when a coolie carries a load in his head, he extracts force along the vertical direction to support the load on his head. Since no distance is covered in the vertical distance of the force applied, the work performed by the coolie is also zero.

Work was done by a Constant Force

Work is said to be a force when the force produces a displacement of a body on which it acts.

Suppose a force \(\vec F\) acts on a body to have a displacement \(\vec s\) then, work done by the force is

$$ W = \vec F . \vec s \dots (i) $$

If \(\theta \) is the angle between \(\vec F \text {and} \vec s\), as shown in the figure, then from equation (i), we have

$$ W = F . s \cos \theta \dots (ii) $$

When displacement is produced in the direction of application of force as shown in the figure, then \(\theta = 0^o \text {or} \: \cos \theta = \cos 0^o = 1\). From equation (ii), we have

$$ W = F . s \dots (iii) $$

Thus work done by a constant force is the product of force and displacement, when the two vectors \(\vec F \text {and} \vec s\) are in the same direction.

Special Cases

1. When \(\theta \) is an acute angle \( (\text {i.e. } \: \theta < 90^o)\), then work done is positive. For example, when a body falls freely under gravitational pull, the work done by gravity is zero.
2. When \(\theta = 90^o, \: \cos 90^o = 0 \) and
   For example, when a cookie travels on a platform with a load on his head, work done by the coolie is zero.
3. When \(90^o < \theta \leq 180^o \) then \(\cos \theta\) is negative and the work done is negative, i.e. work is done against the force.
   For example, when a body slides over a rough surface, the work done by the frictional force is negative.

Units of Work

As work is the product of force and displacement, its unit depends on the unit of force and displacement. In SI-units, unit of work is N m or joule, J. one joule of work is said to be done when a force of one newton displaces a body through one meter.

In CGS –system, unit of work is erg. One erg of work is said to be done when a force of one dyne displaces a body through one centimeter.

Relation between joule and erg is

\begin{align*} 1 J &= 1 N \times 1 m = 10^5 \text {dyne} \times 100 \text {cm} = 10^7 \text {erg} \\ \therefore 1 J &= 10^7 \text {erg} \end{align*}

Dimension of the work is [ML²T^-2].

Work was done by a Variable Force

Let a variable force acting on a body to displace it from A to B in a fixed direction. We can consider the entire displacement from A to B is made up of a large number of infinitesimal displacements. One such displacement is shown in the figure from P to Q. as the displacement PQ = dx is infinitesimally small, we consider that along this displacement, force is constant in magnitude as well as direction.

Small amount of work done in moving the body from P to Q is

$$ dW = F \times dx = (PS)(PQ) = \text {area of strip PQRS} $$

Total work done in moving the body from A to B is given by

$$ W = \sum d W = \sum Fx dx $$

If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit. The sum approaches definite value equal to the area under the curve CD as shown in the figure.

Hence we can write

\begin{align*} W &= \lim _{dx\to 0} = \sum_{XA}^{XB} Fdx &= \sum_{XA}^{XB} \text {area of the strip PQRS} \\ &= \text {Total area under the curve between F and x-axis from} x=x_A \text {to} x= x_B \\ W &= \text {Area ABCDA} \end{align*}