Work Done by a Constant Force
The work done by a constant force is measured as the product of the magnitude of the force and the distance covered by the body in the direction of the force. Let’s first understand what is work.
The term “work” carries a different meaning in everyday language compared to its definition in physics. For instance, a security guard standing at the entrance of an office or a porter carrying a load across a railway platform would be considered as working in everyday terms since they are being paid for their efforts. However, from a physics standpoint, neither of them would be considered to have done any work.
In physics, work is defined as being done when a force applied to an object causes it to move a certain distance in the direction of the force.
Take the security guard at the office gate: they are not causing any object to move, and both the force and displacement are zero, meaning no work is done. Similarly, when a porter carries a load on their head, they exert a force vertically to support it. However, as the movement occurs horizontally, no distance is covered in the direction of the vertical force, resulting in no work being done in the physics sense.
If a force \( \vec{F} \) displaces a body through displacement \( \vec{S} \) in the direction of force, then work done
\[
W = F S \quad \text{…(1)}
\]
where \( F \) and \( S \) are magnitude of the force vector \( \vec{F} \) and the displacement vector \( \vec{S} \) in the direction of force respectively.
In case, force \( \vec{F} \) acts along a direction making angle \( \theta \) with the direction of displacement \( \vec{S} \) as shown in figure, then work done is found by resolving the force \( \vec{F} \) into two rectangular components:
- \( F_x \) along the direction of displacement \( S \), such that
\[
F_x = F \cos \theta
\] - \( F_y \) along perpendicular to displacement \( S \), such that
\[
F_y = F \sin \theta
\]
As the body does not move in the direction of component of force \( F_y \), the component \( F_y \) does no work.
On the other hand, work done by component \( F_x \) is given by
\[
W = \left| \vec{F}_x \right| \left| \vec{S} \right| = (F \cos \theta) S
\]
or
\[
W = FS \cos \theta \quad \text{…(2)}
\]
In figure as shown, \( \vec{F} \) and \( \vec{S} \) are in the same direction, i.e., \( \theta = 0^\circ \). Setting \( \theta = 0^\circ \) in equation (2), we have
\[
W = FS \cos 0^\circ = FS(1) = FS,
\]
which is the same as given by equation (1).
Therefore, equation (2) can be used to obtain work done by a force in all the cases, i.e., when the force may act or may not act in the direction of displacement. Further, as \( \vec{F} \) and \( \vec{S} \) are vectors, equation (2) may be written as
\[
W = \vec{F} \cdot \vec{S}
\]
Hence, work done by a force may be measured as the dot product of the force applied and the displacement caused by the force. As work done is the dot product of the vectors, it is a scalar quantity.
Also, when \( \vec{F} \) and \( \vec{S} \) are perpendicular to each other, then
\[
W = \vec{F} \cdot \vec{S} = FS \cos 90^\circ = FS(0) = 0
\]
Therefore, if a body moves in a direction perpendicular to that in which the force acts, the force does no work.
Further, in the expression for work done \( (W = \vec{F} \cdot \vec{S}) \), it is understood that force \( \vec{F} \) remains constant (both in magnitude and direction) over the whole displacement \( \vec{S} \).
Dimension and Units of Work
The dimensional formula of work is \([ML^2T^{-2}]\). Its absolute unit in SI is joule and is denoted by \(J\). In the cgs system, the absolute unit of work is erg. From the relation
\[
W = FS, \quad (\text{when } \theta = 0^\circ)
\]
\[
1 \text{ joule} = 1 \text{ newton} \times 1 \text{ metre}
\]
Therefore, work done is said to be 1 joule, if a force of 1 newton displaces a body through 1 meter in the direction of force.
Again, from the equation \( W = FS \), we have
\[
1 \text{ erg} = 1 \text{ dyne} \times 1 \text{ cm}
\]
Work done is said to be 1 erg, if a force of 1 dyne displaces a body through 1 cm in the direction of force.
Relation between Joule and Erg
\[
1J = 1N \times 1m = 10^5 \text{ dyne} \times 100 \text{ cm}
\]
or
\[
1J = 10^7 \text{ erg}
\]
The gravitational unit of work in SI is \( \text{kg m} \) and in the cgs system, the unit is \( \text{g cm} \).
Work done is said to be 1 \( \text{kg m} \), if a force of 1 kgf displaces a body through 1 m in the direction of force.
\[
1 \text{kg m} = 1 \text{kgf} \times 1 \text{m} = 9.8J
\]
Work done is said to be 1 \( \text{g cm} \), if a force of 1 gf displaces a body through 1 cm in the direction of force.
\[
1 \text{g cm} = 1 \text{gf} \times 1 \text{cm} = 980 \text{erg}
\]
Example 1: A man weighing 80 kgf carries a stone of weight 20 kgf to the top of the building 30 m high. Calculate the work done by him. Given $ \displaystyle g=9.8m/{{s}^{2}}$.
Solution:
Here weight of the man = 80 kgf, weight of stone = 20 kgf
Force applied to carry the total weight up,
$ \displaystyle \begin{array}{l}F=80+20=100\text{ kgf}\\=100\times 9.8=980\text{ N}\end{array}$
Height through which weight is carried, $ \displaystyle S=30\text{ m}$
Therefore work done is given by,
$ \displaystyle W=FS=980\times 30=29400J$
Example 2: A body constrained to move along z-axis of a co-ordinate system is subject to a constant force
$ \displaystyle \overrightarrow{F}=-\widehat{i}+2\widehat{j}+3\widehat{k}\text{ N}$,
where $ \displaystyle \widehat{i},\widehat{j},\widehat{k}$ are unit vectors along the X, Y, and Z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the Z-axis?
Solution:
Here, $ \displaystyle \overrightarrow{F}=-\widehat{i}+2\widehat{j}+3\widehat{k}\text{ N}$
Since, body moves a distance of 4 m along Z-axis, $ \displaystyle \overrightarrow{S}=4\widehat{k}\text{ m}$
Now,
$ \displaystyle \begin{array}{l}W=\overrightarrow{F}\cdot \overrightarrow{S}\\=\left( {-\widehat{i}+2\widehat{j}+3\widehat{k}\text{ N}} \right)\cdot \left( {4\widehat{k}} \right)\\=-4\left( 0 \right)+8\left( 0 \right)+12\left( 1 \right)=12\text{ J}\end{array}$
Example 3: Calculate the amount of work done in raising a glass of water weighing 0.5 kg through a height of 20 cm. Take $g = 10 \, \text{ms}^{-2}$.
Solution:
Here, $m = 0.5 \, \text{kg}$
$$
h = 20 \, \text{cm} = \frac{20}{100} = 0.2 \, \text{m}
$$
$$
g = 10 \, \text{ms}^{-2}
$$
Work done $= \text{force} \times \text{distance}$
$ \displaystyle W=mg\times h$
$ \displaystyle =0.5\times 10\times \frac{1}{5}=1\text{J}$
Nature of work done in various situations:
The formula for work done is expressed as:
$$ W = \vec{F} \cdot \vec{S} $$
As a result, the work done can be positive, negative, or zero, depending on the angle between $\vec{F}$ and $\vec{S}$. If the angle is acute, the work is positive; if obtuse, it’s negative; and if the angle is $90^\circ$, the work done is zero. Additionally, if either $\vec{F}$ or $\vec{S}$ is zero, no work is done.
(a) Situations where work done is positive:
- When an object falls freely due to gravity, the work done by the gravitational force is positive.
- When a horse pulls a cart on a flat surface, the resulting work is positive.
In these cases, $\vec{F}$ and $\vec{S}$ point in the same direction.
(b) Situations where work done is negative:
- When an object is pushed across a rough surface, the work done by friction (opposing force) is negative.
- When a positive charge is moved towards another positive charge, the work done by the repulsive electrostatic force is negative.
Here, $\vec{F}$ and $\vec{S}$ are in opposite directions.
(c) Situations where work done is zero:
- When a porter walks along a platform carrying a load on his head, the work done by him is zero.
- When an object is moved along a circular path by a string, the tension in the string does no work.
- When a person holds a heavy object without moving, no work is done.
In the first two scenarios, $\vec{F}$ and $\vec{S}$ are perpendicular, while in the third, $\vec{S}$ is zero.
