Obtain the equation of work done by a variable force in one dimension.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) force whose magnitude or direction (or both) changes with position is called a variable force. Constant forces are rare in nature; variable forces are more commonly encountered.
The figure shows a plot of a varying force $F(x)$ in one dimension versus displacement $x$.
If the displacement $\Delta x$ is very small,the force $F(x)$ can be considered approximately constant over this interval. The work done during this small displacement is equal to the area of the small rectangular strip,given by $\Delta W = F(x) \Delta x$.
The total work done is the sum of the areas of all such shaded rectangular strips from the initial position $x_i$ to the final position $x_f$,which is written as:
$W = \sum_{x_i}^{x_f} F(x) \Delta x$
If the displacement $\Delta x$ is allowed to approach zero,the number of terms in the sum increases without limit,and the sum approaches a definite value equal to the area under the curve.
Therefore,the work done over the whole path is:
$W = \lim_{\Delta x \rightarrow 0} \sum_{x_i}^{x_f} F(x) \Delta x$
$W = \int_{x_i}^{x_f} F(x) dx$
Thus,for a varying force,the work done can be expressed as the definite integral of the force with respect to displacement.

Explore More

Similar Questions

The relationship between force $F$ and displacement $x$ is shown in the figure. The work done by the object for a displacement from $x = 1 \ m$ to $x = 5 \ m$ is equal to ... $J$.

$A$ body of mass $200\,g$ is moving along the $XY$ plane. The work performed by the force given by $\vec F = (2x\hat i + y\hat j)$ acting on it when the body is displaced from $(0, 0)$ to $(1, 2)$ will be equal to ............... $unit$.

$A$ body moves from $x = 0$ to $x = x_1$ under the influence of a force $F = cx$. The work done during this process is:

The relationship between the force $F$ and position $x$ of a body is as shown in the figure. The work done in displacing the body from $x = 1 \text{ m}$ to $x = 5 \text{ m}$ will be ........ $J$.

$A$ particle is projected with velocity $v_{0}$ along the $x-$axis. $A$ damping force is acting on the particle which is proportional to the square of the distance from the origin,i.e.,$ma = -\alpha x^{2}$. The distance at which the particle stops is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo