Work is said to be done if a force acting on a body displaces it through some distance in the direction of applied force. Energy and power of a body is associated with the amount of work done by the body.
Let us consider a force $F$ displaces a body having mass $m$ through displacement $s$ in horizontal direction in which angle between the direction of motion and application of force be $θ$ as shown in figure. Then, this force has two components; $F\cos θ$ and $F\sin θ$ in which $F\cos θ$ is responsible for the displacement.
Work done $(W)$ $=$ Component of force along the direction of motion $×$ displacement
\[W = F\cos θ × s\] \[W = Fs\cos θ\] This is the work done by constant force. In vector form, \[W=\vec{F}\vec{s}\]
Case I: When direction of motion and the direction of force is same $(θ = 0°)$ then, \[W = Fs \]
Case II: When force and direction of motion are in opposite direction $(θ = 180°)$ then, \[W = -Fs\]
Case III: When force and direction of motion are perpendicular with each other $(θ = 90°)$ then, \[W = 0\]
Graphically, work can be obtained from the area between the force and displacement axis as shown in figure.
Work is an algebraic quantity having no definite direction. It is a scalar quantity.
In SI system, the unit of force is Newton and the unit of distance covered by the body is metre. Hence, the unit of work done is Newton metre $(\text{Nm})$, also known as Joule $(J)$. In CGS system, the unit of force is dyne and the unit of distance covered is cm, and the unit of work is Erg. \[1 \text{ Joule}=10^7 \text{ Erg}\]
Work Done Due To Variable Forces
Let us consider a body is covering distance by the application of the variable force. Suppose the variable force displaces the body from $A$ to $B$ as shown in graph.
This displacement $AB$ can be considered to be made up of large number of infinitesimal displacements. In such a small displacement, force is almost constant. Let $dx$ from $P$ to $Q$ be the small displacement made by the constant force $F$. Then, small amount of work done in moving the body from infinitesimally small distance $dx$ is, \[dW = Fdx = QS × PQ = \text{Area of PQSR}\]
Total work done in moving the body from $A$ to $B$ is, \[W=\sum dW\] \[W=\lim_{dx\to 0}\sum F\;dx\] \[W=\int_{x_1}^{x_2} Fdx\] \[W=\int_{x_1}^{x_2} \text{Area of PQSR}\]
$W=$ Total area under the curve between force and displacement axis from $x=x_1$ to $x_2$ \[W=\text{Area of ABCD}\]
Thus, the area between the force and the displacement axis gives the work done by variable force.
Q. An object is pulled along the ground by a 75 N force directed 30° above the horizon. How much work is done by the force in pulling the object through 8√3 m?
\[\text{Work done}=\text{Force}×\text{Displacement}\] \[=(F\cos 30)×8√3\] \[=75\cos 30×8√3\] \[=900 \text{ } J\]
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