Work Done by a Variable Force: Elaboration, Formula, Examples

Work Done by a Variable Force: All natural forces may be divided into two categories. There are two types of forces: constant forces and variable forces. Did you realise that the forces we experience on a daily basis are largely changeable in nature? Spring force, electrostatic force, gravitational field force, magnetic force, and other variable forces are examples. When force is applied in the direction of the force, there is a displacement in the system, which is called forced work. By splitting displacement into tiny intervals, the work done by a variable force may be calculated. Integration is required to compute the work done in the situation of a variable force.

In this article, we will learn about a few different types of variable force; we will also learn how to calculate the work done by them.

Types of Forces

1. Constant force:- The force whose magnitude and direction do not change throughout the motion of the body is called constant force. Examples of constant forces are kinetic friction force, hydrostatic force, weight, etc.
2. Variable force:- The force whose magnitude and direction changes during the motion of the body is called Variable force. Examples of variable forces are spring force, Electrostatic force, Magnetic force, etc.

Work Done by a Force

a) Work done by constant force:- In this case, the force will be constant throughout the displacement. Calculation of work done by constant forces is simple. The work done by constant force can be directly calculated by calculating the dot product of force and displacement vector.

The magnitude of work done,
\(W = \overrightarrow F \cdot \overrightarrow d = Fd\cos \theta \)
b) Work done by variable force:- Calculation of the work done by the variable force is a bit more complex than calculation of work done by a constant force. As the force always keeps on changing in the case of variable force, the dot product is calculated for elemental displacement. During this elemental change in displacement, the force can be treated as constant. When the integration is performed on the whole displacement, we will the total work done by the variable force. For example, in the case of a stretching and compressing of spring, bringing two charge particles close, etc.
\({\rm{d}}W = \overrightarrow F \cdot {\rm{d}}\overrightarrow s \)
\(W = \int {\overrightarrow F \cdot {\rm{d}}\overrightarrow s } \)
To understand the calculation of magnitude by a variable force, let us take an example of any general force doing work. The force is given by, \(\overrightarrow F = \left( {m{x^2} + nx} \right)\widehat i.\) Let us calculate the work done by this force in moving from \(x = a\) to \(x = b.\)
Let the elemental displacement occurred due to the application of force at any instant be \(\left( {{\rm{d}}x} \right)\widehat i\). Then the differential work done will be,
\({\rm{d}}W = \overrightarrow F \cdot {\rm{d}}\overrightarrow s \)
\( \Rightarrow {\rm{d}}W = \left( {m{x^2} + nx} \right)\widehat i \cdot \left( {\rm{d}}{x} \right)\widehat i\)
\( \Rightarrow {\rm{d}}W = \left( {m{x^2} + nx} \right){\rm{d}}x\)
\( \Rightarrow W = \int {\rm{d}}{W} = \int_a^b {\left( {m{x^2} + nx} \right){\rm{d}}x} \)
\( = m\left( {\frac{{{b^3} – {a^3}}}{3}} \right) + n\left( {\frac{{{b^2} – {a^2}}}{2}} \right).\)

Calculation of Work Done from Graph

If the graph is plotted between the force acting on the object and displacement of the object, then the area under the curve will give the work done by the force.

Fig-a

Fig-b

From fig-a, we can observe that the graph of constant force is a straight line. But in the case of variable force (Fig-b), the graph is in the form of a curve. Work done by a variable force is determined by dividing displacement into small intervals. The shaded region in both cases gives work done.

Work Done by an External Force During Elongation of a Spring

When the spring has a natural length, it will not exert any force. But when it is elongated or compressed from its natural length, it will exert force. We know that the spring force is proportional to the displacement of the object from the equilibrium position (natural length). Hence, the force acting at each instant during the compression and extension of the spring will be varying with displacement from natural length.
Spring force, \(F = – kx\)
Where,
\(k\) is the spring constant,
\(x\) is the displacement from the natural length of the spring.

Since the force acting on the block changes at each position, the dot product is calculated for elemental displacement. This elemental work is then integrated over total displacement.
Let us calculate the work done by external force in elongating a spring by \(x_m\) length from its natural length, as shown in the above figure. We know that the spring force at displacement \(x\) from natural length will be,
\(F_x = -kx\)
For the block to be in equilibrium, the external force of equal magnitude will act in the opposite direction. So, the magnitude of external force at displacement \(x\) from natural length will be,
\(F_s = kx\)
The differential work done by external force will be,
\({\rm{d}}W = \overrightarrow F \cdot {\rm{d}}\overrightarrow s = (kx){\rm{d}}x\)
Total work done by external force will be,
\(W = \int_0^{{x_m}} {{\text{d}}W} = \int_0^{{x_m}} {(kx} ){\rm{d}}x\)
\( = \left[ {\frac{{k{x^2}}}{2}} \right]_0^{{x_m}}\)
\( = \frac{{kx_m^2}}{2}\)
So, the work done by external force in elongating a spring by \(x_m\) unit will be \(W = \frac{{kx_m^2}}{2}.\) The same amount of work will be done in case of compression also.
But the work done by the spring force will be negative of work done by the external force.
So, the work done by spring force will be \({W_s} = – \frac{{kx_m^2}}{2}.\)

Sample Problems

Q.1. Mass \((m)\) of an object is \(2\;\rm{kg}.\) This object moves due to variable force in the direction of the positive \(x-\)axis. Force variation is given by the function, \(\;\overrightarrow F = \left( {3{\text{ }} + {\text{ }}2x} \right)\widehat i\rm{N}.\) Determine the work done when object moves from \(x = 0\) to \(x = 5.\)
Ans: As the magnitude of the force is varying, we will have to calculate differential work and then integration over total displacement.
At any displacement \(x\), the magnitude of force, \(\overrightarrow F = \left( {3 + 2x} \right)\widehat i\,{\text{N}}\)
Let the differential displacement be \(\left( {{\rm{d}}x} \right)\widehat i.\)
\({\rm{d}}W = \overrightarrow F \cdot {\rm{d}}\overrightarrow s \)
\(W = \int_0^5 {{\rm{d}}W = \int_0^5 {\left( {3 + 2x} \right)} {\rm{d}}x} \)
\( = \int_0^5 {\left( {3 + 2x} \right){\rm{d}}x} \)
\( = \left[ {3x + \frac{{2{x^2}}}{2}} \right]_0^5\)
\(= 30\;\rm{J}\)
Thus, the total work done will be equal to \(30\;\rm{J}.\)

Q.2. A \(40\,\rm{m}\) long cable that weighs \(5\,\rm{kg}\,\rm{m}^{-1}\) is hanging from the roof of a very tall building. How much work is required to lift it all to the roof level? (Take \(g = 9.8\,\rm{m}\,\rm{s}^{-2}\))
Ans: Suppose that \(x\) metres of the cable has already been lifted to roof level so that \((40 – x)\) metres still hanging from the roof.
The weight of the remaining chain will be \(5(40 − x)\,\rm{kg.}\)
So, the force requiring to pull elemental length dx of the chain will be,
\(F = mg = 9.8 \times 5(40 – x) = 49(40 – x)\,{\text{N}}\)
The work required to lift this remaining cable by distance \({\rm{d}}x\) will be,
\({\rm{d}}W = 49\left( {40 – x} \right){\rm{d}}x\,{\text{J}}\)
So, the total work done will be,
\(W = \int_0^{40} {\rm{d}}W = \int_0^{40} {49\left( {40 – x} \right){\rm{d}}x\,{\text{J}}} \)
Evaluating the definite integral, we get \(W\)
\(W = \left[ {49\left( {40 – \frac{{{x^2}}}{2}} \right)} \right]_0^{40}\,{\text{J}}\)
\( = 49\left[ {40\left( {40} \right) – \frac{{{{\left( {40} \right)}^2}}}{2}} \right]\)
\( = 49 \times \left( {800} \right) = 39200\,{\text{J}}\)
So, the total work needed will be equal to \(39200\,{\text{J}}.\)

Summary

Variable force occurs when the direction and amount of a force vary throughout the motion of a body. Magnetic force, spring force, and electrostatic force are examples of variable forces. The majority of the forces we experience in our daily lives are variable forces. By splitting displacement into tiny intervals, the work done by a variable force may be computed.

When any force displaces any object and does work on it, the force may change its direction, magnitude or both. The forces of these kinds are called variable forces. Calculation of the work done by a variable force is a bit more complex than the calculation of work done by a constant force. Work done by a variable force is determined by dividing displacement into small intervals. Differential work done in this small interval is calculated and integrated over the whole displacement to get total work.

\({\rm{d}}W = \overrightarrow F \cdot {\rm{d}}\overrightarrow s \)

\(W = \int {{\rm{d}}W} = \int {\overrightarrow F \cdot } {\rm{d}}\overrightarrow s \)

The work done by force can also be calculated from the graphical method. The area under the curve in the graph of force vs displacement will give the magnitude of work done by the force.

FAQs on Work Done by Variable Force

Q.1. What is meant by variable force?
Ans: Force is a vector quantity. A force is said to be a variable force if magnitude or direction or both changes with time.

Q.2. Give some examples of variable force?
Ans: In our daily life, we observe many different variable forces. Some of them are spring force, magnetic force, Electrostatic force, Drag force during fall of the ball, Pressure force experienced by scuba divers, etc.

Q.3. What does the graph of force vs displacement represent?
Ans: The graph of force vs displacement tells us the variation of force acting on the body at different intervals of time. The area under the graph represents the work done during the displacement.

Q.4. How do you calculate the work done by a variable force?
Ans: Work done by a variable force is determined by dividing displacement into small interval. Differential work is calculated during this small displacement. We then integrate over total displacement to get total work.
\(W = \int {{\rm{d}}W = \int {\overrightarrow F } \cdot \overrightarrow {{\text{d}}r} } \)
\(W = \int {F{\text{d}}} r\cos \left( \theta \right)\)
Here, \(\theta\) is the angle between the force and differential displacement \({\text{d}}r.\)

Q.5. What is the dimensional formula of work?
Ans: We know that work done by any force is given by, \(W = \overrightarrow F \cdot \overline s .\)
Thus, \(\left[ W \right] = \left[ F \right]\left[ s \right]\)
\( = \left[ m \right]\left[ a \right]\left[ d \right]\)
\( = ML{T^{ – 2}}L = M{L^2}{T^{ – 2}}.\)

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