Work Done By Variable Force

When a force acts on an object and causes displacement, work is done. In the real world, forces often vary rather than remain constant, leading to what is known as variable force. Understanding the work done by a variable force is crucial in fields such as physics and engineering, where it helps to explain how energy is transferred in various situations. For example, when a car accelerates, the engine exerts a variable force on the car, changing its speed and kinetic energy. Similarly, when you stretch a spring or lift an object using a pulley system, the force exerted changes with distance, making the work done dependent on how the force varies with displacement. This concept is vital for designing efficient machines, optimizing energy usage, and predicting system behaviour in complex environments

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

$W=\int \vec{F} \cdot \overrightarrow{d s}$

Where $\vec{F}$ is a variable force and $\overrightarrow{d s}$ is a small displacement

When Force is Time-Dependent

And we can write $d \vec{s}=\vec{v} d t$

So,

$
W=\int \vec{F} \cdot \vec{v} d t
$
Where $\vec{F}$ and $\vec{v}$ 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) $\frac{1}{2}\left(a L^2+b L^3\right)$
3) $\frac{a L^2}{2}+\frac{b L^3}{3}$
4) $\frac{1}{2}\left(\frac{a L^2}{2}+\frac{b L^3}{3}\right)$

Solution:

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

wherein
$\vec{F}$ is variable force and $\overrightarrow{d s}$ 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{aligned}
& W=\int_0^L\left(a x+b x^2\right) d x \\
& =a \int_0^L x d x+b \int_0^L x^2 d x \\
& w=\frac{a L^2}{2}+\frac{b L^3}{3}
\end{aligned}
$

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 \vec{F} \cdot \overrightarrow{d s}
$

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

$
\begin{aligned}
& V=\frac{d x}{d t}=6 t \\
& \begin{aligned}
\text { Work done } & =\frac{1}{2} m v^2-0 \\
& =\frac{1}{2}(2)(6 \times 5)^2-0 \\
& =900 \mathrm{~J}
\end{aligned}
\end{aligned}
$

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 $F=6 t$

$
\begin{aligned}
& \mathrm{F}=\mathrm{ma}=m \frac{d v}{d t}=6 \mathrm{t} \\
& \frac{d v}{d t}=6 t_{(\text {Since } \mathrm{m}=1 \mathrm{~kg})} \\
& \underline{\mathrm{dv}}=6 \mathrm{t} \mathrm{dt}
\end{aligned}
$
On integrating, $\int_0^v d V=6 \int_0^1 t d t=3$
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=\Delta K E=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:

$\begin{aligned} & F=200 \mathrm{~N} \quad \text { for } 0 \leq \mathrm{x} \leq 15 \\ & =200-\frac{100}{15}(\mathrm{x}-15) \text { for } \quad 15 \leq \mathrm{x}<30 \\ & \mathrm{~W}=\int \mathrm{Fdx} \\ & =\int_0^{15} 200 \mathrm{dx}+\int_{15}^{30}\left(300-\frac{100}{15} \mathrm{x}\right) \mathrm{dx} \\ & =200 \times 15+300 \times 15-\frac{100}{15} \times \frac{\left(30^2-15^2\right)}{2} \\ & =3000+4500-2250 \\ & =5250 \mathrm{~J}\end{aligned}$

Hence, the answer is the option (4).

Example 5: A particle experiences a variable force $\vec{F}=\left(4 \mathrm{x} \hat{\mathrm{i}}+3 \mathrm{y}^2 \hat{\mathrm{j}}\right)$ 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,

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

$
\begin{aligned}
& \begin{aligned}
\bar{F} & =F_x \hat{i}+F_y \hat{j} \\
\mathrm{~F}_{\mathrm{x}} & =4 \mathrm{x}, \quad \mathrm{F}_{\mathrm{y}}=3 \mathrm{y}^2 \\
\mathrm{dW} & =\bar{F} \cdot d \bar{s} \\
& =\left(F_x \hat{i}+F_y \hat{j}\right) \cdot\left(d_x \hat{i}+d_y \hat{j}\right) \\
\mathrm{dW} & =\mathrm{F}_{\mathrm{x}} \mathrm{dx}+\mathrm{F}_{\mathrm{y}} \mathrm{dy}
\end{aligned} \\
& \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}=\Delta \mathrm{KE}
\end{aligned}
$

(from Work-energy theorem)

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

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.