Work Done By Variable Force

Work Done By Variable Force

Force is a vector quantity. So it has a magnitude as well as direction. A variable force means when its magnitude its direction or both varies with position.

And work done by the variable force is given by

$W=\int \overrightarrow{F}\cdot \overrightarrow{ds}$

Where $\overrightarrow{F}$ is a variable force and $\overrightarrow{ds}$ is a small displacement

When Force is Time-Dependent

And we can write $d\overrightarrow{s}=\overrightarrow{v}dt$

So,

$W=\int \overrightarrow{F}\cdot \overrightarrow{v}dt$
Where $\overrightarrow{F}$ and $\overrightarrow{v}$ are force and velocity vectors at any instant.

Work Done Calculation by Force Displacement Graph

The area under the force-displacement curve with the proper algebraic sign represents work done by the force.

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Solved Examples Based on Work Done By Variable Force

Example 1: When a rubber band is stretched by a distance $x$, it exerts a restoring force of magnitude $\mathrm{F}=\mathrm{ax}+{\mathrm{bx}}^2$ where a and b are constants. The work done in stretching the unstretched rubber band by L is :

1) $a{L}^2+b{L}^3$
2) $\frac{1}{2}(a{L}^2+b{L}^3)$
3) $\frac{a{L}^2}{2}+\frac{b{L}^3}{3}$
4) $\frac{1}{2}(\frac{a{L}^2}{2}+\frac{b{L}^3}{3})$

Solution:

Definition of work done by a variable force
$W=\int \overrightarrow{F}\cdot \overrightarrow{ds}$

wherein
$\overrightarrow{F}$ is variable force and $\overrightarrow{ds}$ is a small displacement
At $x=x$

$\mathrm{F}=\mathrm{ax}+b{x}^2$
Work done in displacing rubber through $\mathrm{dx}=\mathrm{Fdx}$

$\begin{array}{rl}& W={\int }_0^L(ax+b{x}^2)dx\\ & =a{\int }_0^{L}xdx+b{\int }_0^L{x}^{2}dx\\ & w=\frac{a{L}^2}{2}+\frac{b{L}^3}{3}\end{array}$

Hence, the answer is the option (3).

Example 2: A force acts on a 2kg object so that its position is given as a function of time as x = 3t² + 5. What is the work (in Joule) done by this force in the first 5 seconds?

1) 850

2) 875

3) 950

4) 900

Solution:

Definition of work done by a variable force
$W=\int \overrightarrow{F}\cdot \overrightarrow{ds}$

wherein
$\overrightarrow{F}$ is variable force and $\overrightarrow{ds}$ is a small displacement
Work done $=$ change in kinetic energy

$\begin{array}{rl}& V=\frac{dx}{dt}=6t\\ & \begin{array}{rl}\text{Work done}& =\frac{1}{2}m{v}^2-0\\ & =\frac{1}{2}(2)(6\times 5{)}^2-0\\ & =900\mathrm{J}\end{array}\end{array}$

Hence, the answer is the option (4).

Example 3: A time-dependent force F=6t acts on a particle of mass 1 kg. If the particle starts from rest, the work (in Joule) is done by the force during the first 1 sec. will be :

1) 4.5

2) 22

3) 9

4) 18

Solution:

Force is given $F=6t$

$\begin{array}{rl}& \mathrm{F}=\mathrm{ma}=m\frac{dv}{dt}=6\mathrm{t}\\ & \frac{dv}{dt}=6t_{(\text{Since}\mathrm{m}=1\mathrm{kg})}\\ & \underset{―}{\mathrm{dv}}=6\mathrm{t}\mathrm{dt}\end{array}$
On integrating, ${\int }_0^{v}dV=6{\int }_0^{1}tdt=3$
Therefore $\mathrm{v}=3\mathrm{m}/\mathrm{s}$
Therefore change in kinetic energy in one second $=\frac{1}{2}\times m\times 3^2-0=4.5\mathrm{J}$
Since the change in kinetic energy is equal to the work done.

$\therefore W=\mathrm{\Delta }KE=4.5\mathrm{J}$

Hence, the answer is the option (1).

Example 4: A person pushes a box on a rough horizontal surface. He applies a force of 200N over a distance of 15m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100N. The total distance through which the box has been moved is 30m. What is the work done by the person during the total movement of the box?

1) 3280 J

2) 2780 J

3) 5960 J

4) 5250 J

Solution:

$\begin{array}{rl}& F=200\mathrm{N}\text{for}0\le \mathrm{x}\le 15\\ & =200-\frac{100}{15}(\mathrm{x}-15)\text{for}15\le \mathrm{x}<30\\ & \mathrm{W}=\int \mathrm{Fdx}\\ & ={\int }_0^{15}200\mathrm{dx}+{\int }_{15}^{30}(300-\frac{100}{15}\mathrm{x})\mathrm{dx}\\ & =200\times 15+300\times 15-\frac{100}{15}\times \frac{({30}^2-{15}^2)}{2}\\ & =3000+4500-2250\\ & =5250\mathrm{J}\end{array}$

Hence, the answer is the option (4).

Example 5: A particle experiences a variable force $\overrightarrow{F}=(4\mathrm{x}\hat{\mathrm{i}}+3{\mathrm{y}}^2\hat{\mathrm{j}})$ in a horizontal $\mathrm{x}-\mathrm{y}$ plane. Assume distance in meters and force is Newton. If the particle moves from point $(1,2)$ to point $(2,3)$ in the $\mathrm{x}-\mathrm{y}$ plane, then the Kinetic energy changes by :

1) 50.0 J
2) 12.5 J
3) 25.0 J
4) 0 J

Solution:

The force experienced by particles in a horizontal $X-Y$ plane is,

$\overline{F}=4x(\hat{i})+3y^2(\hat{j})$
Comparing with,

$\begin{array}{rl}& \begin{array}{rl}\overline{F}& =F_x\hat{i}+F_y\hat{j}\\ {\mathrm{F}}_{\mathrm{x}}& =4\mathrm{x},{\mathrm{F}}_{\mathrm{y}}=3{\mathrm{y}}^2\\ \mathrm{dW}& =\overline{F}\cdot d\overline{s}\\ & =(F_x\hat{i}+F_y\hat{j})\cdot (d_x\hat{i}+d_y\hat{j})\\ \mathrm{dW}& ={\mathrm{F}}_{\mathrm{x}}\mathrm{dx}+{\mathrm{F}}_{\mathrm{y}}\mathrm{dy}\end{array}\\ & \mathrm{W}=\int \mathrm{dW}={\int }_{\mathrm{x}=1}^{\mathrm{x}=2}4\mathrm{x}\mathrm{dx}+{\int }_{\mathrm{y}=2}^{\mathrm{y}=3}3{\mathrm{y}}^2\mathrm{dy}=\mathrm{\Delta }\mathrm{KE}\end{array}$

(from Work-energy theorem)

$\begin{array}{rl}& \mathrm{W}=4{\left[\frac{{\mathrm{x}}^2}{2}\right]}_1^2+3{\left[\frac{{\mathrm{y}}^3}{3}\right]}_2^3\\ & =4[2-\frac{1}{2}]+3[9-\frac{8}{3}]\\ & \mathrm{W}=6+19=25\mathrm{J}=\mathrm{\Delta }\mathrm{KE}\end{array}$

Hence, the answer is the option (3)

Summary

The article discusses the concept of work done by variable forces, which are forces whose magnitude, direction, or both change with position. It explains how work done by such forces can be calculated using integration and how it can be visualized using force-displacement graphs. The article also provides several solved examples to illustrate the application of these concepts, such as calculating work done by stretching a rubber band, analyzing time-dependent forces, and determining the energy changes in particles experiencing variable forces. These examples help in understanding the practical implications of variable force in real-life situations.