Work Done by a Variable Force
A force applied to a body may not always be constant. It can be a varying force that keeps on changing. We can, therefore, calculate work done by a variable force.
Consider a body subjected to a variable force \( F(x) \), and assume this force is acting along the fixed direction, say the x-axis, but with varying magnitude, as shown in the figure.
We need to calculate the work done in moving the body from point \( A \) to point \( B \) under the influence of this varying force. To do this, we assume that the entire displacement from \( A \) to \( B \) is composed of a large number of infinitesimally small displacements. One such displacement is shown in the figure as \( P \) to \( Q \).
As the displacement \( PQ = dx \) is infinitesimally small, we consider that along this displacement, the force is constant in magnitude \( = (PS) \) as well as in direction.
Therefore, small amount of work done in moving the body from \( P \) to \( Q \) is:
$$ dW = F(x) \cdot dx = (PS) \cdot (PQ) $$
Where \( W \) is the area of strip \( P \) to \( Q \).
Total work done in moving the body from \( A \) to \( B \) is given by:
$$ W = \sum dW $$
$$ W = \sum F(x) \cdot dx $$
If the displacements are allowed to approach zero, then the number of terms in the summation increases without limit. And the sum approaches a definite value equal to the area under the curve \( CD \) as shown in the figure.
Hence, we may rewrite:
$$ W = \lim_{dx \to 0} \sum F(x) \cdot dx $$
or
Using the concepts from integral calculus, we can express work as:
$$ W = \int_{x_A}^{x_B} F(x) \, dx $$
Where:
– \( x_A \) is the coordinate OA.
– \( x_B \) is the coordinate OB.
In this context, \( W \) represents the area under the strip PQRS.
Therefore, the total work done is the area under the curve between the force \( F \) and the x-axis, from \( x_A \) to \( x_B \).
This gives us:
$$ W = \text{Area of region ABCDA} $$
Thus, the work done by a variable force is numerically equal to the area beneath the force-displacement curve.
Important: When calculating the area under the force-displacement curve, remember to assign the correct sign to determine the net work done.
Detailed Mathematical Approach (for work done by a variable force):
Let’s consider a situation where you need to calculate the work done when moving an object from point \( A \) (at position \( S_A \)) to point \( B \) (at position \( S_B \)) under the influence of a varying force, as depicted in figure below. Here, \( S_A \) and \( S_B \) represent the distances from a reference point to points \( A \) and \( B \) respectively.
At any point during the movement, assume the object is at position \( P \), where the force acting on it is \( F \). As the object undergoes an infinitesimal displacement \( PQ = ds \), the small amount of work done in moving the object from \( P \) to \( Q \) is given by:
$$ dW = F \cdot ds = (PS) \cdot (PQ) $$
The total work done in moving the object from \( A \) to \( B \) can be represented as:
$$ W = \sum dW $$
$$ W = \sum F \cdot ds $$
If the displacement increments \( ds \) become infinitesimally small, the number of terms in the summation increases indefinitely, leading to the sum approaching a specific value equivalent to the area beneath the curve \( CD \), as shown in the figure.
We can therefore rewrite the expression as:
$$ W = \lim_{ds \to 0} \sum F(s) \, ds $$
Or:
$$ W = \int_{S_A}^{S_B} F(s) \, ds $$
For more precision, if \( d\vec{s} \) becomes very small, the total work done in moving the object from \( A \) to \( B \) can be obtained by integrating:
$$ W = \int_{S_A}^{S_B} \vec{F} \cdot d\vec{s} $$
Example: When considering a block attached to a spring, the force acting on the block is proportional to the spring’s elongation, which is not constant. As the block moves, the elongation changes, resulting in a variable force. This is an instance of work done by a variable force.
Key Points:
- The total work done in displacing a body under multiple forces equals the work done by the resultant force. If the body is in equilibrium (either static or dynamic), the resultant force \( R = 0 \), hence \( W = 0 \).
- When the force is always perpendicular to the direction of motion, the work done is zero. In other words, if the force and displacement are at right angles, the work done, such as by the centripetal force, is always zero.
- Work done is also influenced by the chosen frame of reference.
Do you know?
Positive work on any system implies that energy is being transferred to the system, whereas negative work implies a transfer of energy from the system.
Example 1: A force \( F = (10 + 0.5x) \) acts on a particle moving along a straight line, where \( F \) is in newtons and \( x \) is in meters. Find the work done by this force as the particle moves from \( x_1 = 0 \) to \( x_2 = 2 \) meters.
Solution:
Given that \( F = (10 + 0.5x) \), the work done by this force over a small displacement \( dx \) is:
$$ dW = F \cdot dx = (10 + 0.5x)dx $$
Total work done,
$ \displaystyle W=\int\limits_{{x=0}}^{{x=2}}{{\left( {10+0.5x} \right)dx}}$
$ \displaystyle \begin{array}{l}W=\left[ {10x+0.5\frac{{{{x}^{2}}}}{2}} \right]\\=10\left( {2-0} \right)+\frac{{0.5}}{2}\left( {{{2}^{2}}-0} \right)\\=20+1=21\text{ J}\end{array}$
