Physics > Work Energy and Power > 2.0 Work done by a constant force
Work Energy and Power
1.0 Introduction
2.0 Work done by a constant force
3.0 Spring Force
4.0 Conservative & Non-conservative forces
5.0 Kinetic Energy $(K)$
6.0 Potential energy $\left( {\Delta U} \right)$
6.1 Potential energy $\left( {\Delta U} \right)$ is negative of the work done by conservative forces.
6.2 Types of potential energy
6.3 Law of conservation of mechanical energy
7.0 Work energy theorem
8.0 Power
9.0 Types of equilibrium
10.0 Work done by a distributed mass
2.2 Work done by a variable force
6.2 Types of potential energy
6.3 Law of conservation of mechanical energy
Let us consider a variable force $F$ acting on a body. The force $F$ is a function of $x$ as shown by the graphical relation.
Work done on a body by a variable force can be written as, $$ W = \int\limits_{{x_1}}^{{x_2}} {\overrightarrow F (x).d\overrightarrow x } $$
The area marked by brown color is the work done by a variable force in displacing a body by a distance $dx$.
The area marked by sky blue color is the work done by a variable force in displacing the body from $x_1$ to $x_2$.
Question 5. A variable force $F=3x^2+2x$ is applied on a body. Find the work by the force in displacing the body along the $x$ axis by a distance $2m$ from the origin.
Solution: Work done is given as $$\begin{equation} \begin{aligned} W = \int {\overrightarrow F .d\overrightarrow x } \\ W = \int\limits_0^2 {\left( {3{x^2} + 2x} \right)dx} \\ W = \left[ {{x^3} + {x^2}} \right]_0^2 \\ W = 12J \\\end{aligned} \end{equation} $$
Question 6. A force $\overrightarrow F = \left( {3{x^2}\widehat i + 2y\widehat j + 2\widehat k} \right)$ displaces a body from position $(2,3,4)$ to $(1,2,3)$. Find the work done.
Solution: Small displacement in 3D space is given as, $d\overrightarrow r = \left( {dx\widehat i + dy\widehat j + dz\widehat k} \right)$.
Therefore the work done is, $$\begin{equation} \begin{aligned} W = \int\limits_{(2,3,4)}^{(1,2,3)} {\overrightarrow F .d\overrightarrow r } \\ W = \int\limits_{(2,3,4)}^{(1,2,3)} {\left( {3{x^2}\widehat i + 2y\widehat j + 2\widehat k} \right).\left( {dx\widehat i + dy\widehat y + dz\widehat k} \right)} \\ W = \int\limits_{(2,3,4)}^{(1,2,3)} {\left( {3{x^2}dx + 2ydy + 2dz} \right)} \\ W = \left[ {{x^3}} \right]_2^1 + \left[ {{y^2}} \right]_3^2 + \left[ {2z} \right]_4^3 \\ W = (1 - 8) + (2 - 9) + (6- 8) \\ W = - 16J \\\end{aligned} \end{equation} $$
Note: Work done is negative because the direction of force and displacement is not same.
