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Physics · Ch 5 — Work, Energy and Power

Work Done by a Variable Force

5.5

Work Done by a Variable Force

Work Done by a Variable Force

In real life, forces are rarely constant. When you stretch a spring, the force you apply increases the further you pull. When a rocket lifts off, the thrust changes as fuel burns. The simple formula W=Fdcos⁡θW = F d \cos \theta only works when the force FF stays the same throughout the displacement. For a force that changes with position, we need a different approach.

The Basic Idea: Breaking the Motion into Tiny Steps

Imagine a particle moving along the xx-axis from xix_i to xfx_f, acted upon by a force F(x)F(x) that depends on the position xx. The force is different at every point along the path.

We can handle this by dividing the total displacement into a large number of very small intervals, each of width Δx\Delta x. Over such a tiny interval, the force is approximately constant. If we take the interval from xx to x+Δxx + \Delta x, the work done over that small step is approximately:

ΔW≈F(x) Δx\Delta W \approx F(x) \, \Delta x

The total work done over the entire displacement is the sum of the work done over all these tiny intervals:

W≈∑xixfF(x) ΔxW \approx \sum_{x_i}^{x_f} F(x) \, \Delta x

This approximation becomes exact as we make the intervals infinitesimally small — that is, as Δx→0\Delta x \to 0. In this limit, the sum becomes an integral.

W=∫xixfF(x) dxW = \int_{x_i}^{x_f} F(x) \, dx

This is the fundamental definition of work done by a variable force in one dimension. The integral literally adds up the force at every point along the path, multiplied by the infinitesimal displacement dxdx.

Geometric Interpretation

The integral ∫xixfF(x) dx\int_{x_i}^{x_f} F(x) \, dx has a clear geometric meaning: it is the area under the curve of F(x)F(x) versus xx, between the limits xix_i and xfx_f.

If you plot force on the vertical axis and position on the horizontal axis, the work done is the area bounded by the curve, the xx-axis, and the vertical lines at x=xix = x_i and x=xfx = x_f.

Watch out

The area counts as positive when the force and displacement are in the same direction (the curve lies above the xx-axis). If the force opposes the motion (the curve dips below the xx-axis), the area contributes negative work. The total work is the net signed area — not the total area.

A Concrete Example: Stretching a Spring

Consider a spring that obeys Hooke's law. The force required to stretch or compress the spring by a distance xx from its natural length is:

F(x)=−kxF(x) = -k x

Here kk is the spring constant (a measure of the spring's stiffness), and the negative sign indicates that the spring force always opposes the displacement — it's a restoring force.

If we want to calculate the work done by an external agent (like your hand) in stretching the spring slowly from x=0x = 0 to x=xmx = x_m, we need the force that the external agent applies. To stretch the spring at constant speed (so that kinetic energy doesn't change), the external force must exactly balance the spring force at every point:

Fext(x)=+kxF_{\text{ext}}(x) = +k x

The work done by this external force is:

Wext=∫0xmFext(x) dx=∫0xmkx dxW_{\text{ext}} = \int_{0}^{x_m} F_{\text{ext}}(x) \, dx = \int_{0}^{x_m} k x \, dx

Evaluating the integral:

Wext=k[x22]0xm=12kxm2W_{\text{ext}} = k \left[ \frac{x^2}{2} \right]_{0}^{x_m} = \frac{1}{2} k x_m^2 …

Figure 5.3.a(a) The shaded rectangle represents the work done by the varying force F(x) over the small displacement Δx, ΔW = F(x) Δx.
Fig. 5.3.a — (a) The shaded rectangle represents the work done by the varying force F(x) over the small displacement Δx, ΔW = F(x) Δx.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure shows a graph with force F(x)F(x) plotted on the vertical axis and displacement xx on the horizontal axis. The curve itself rises from the origin, then gradually flattens out — it represents a force that changes with position, not a constant one. Two vertical lines are drawn at xix_i and xfx_f, marking the start and end of the motion we care about.

The key visual is this: the area under the curve between xix_i and xfx_f is approximated by a series of tall, thin rectangles. Each rectangle has width Δx\Delta x and height equal to the force at that particular xx — so its area is F(x)ΔxF(x) \Delta x. One of these rectangles is shaded, and next to it the figure labels its area as ΔA=F(x)Δx\Delta A = F(x) \Delta x.

What the figure is teaching is the fundamental idea of integration before you have the integral sign. When the force is not constant, you cannot just multiply FF by total displacement. Instead, you break the total displacement into many tiny steps Δx\Delta x. Over each tiny step, the force is approximately constant, so the work done in that step is ΔW≈F(x)Δx\Delta W \approx F(x) \Delta x — exactly the area of that thin rectangle. The total work is then the sum of the areas of all these rectangles.

Important

The physical meaning of the shaded rectangle: it is the work done by the force over a single small displacement Δx\Delta x at a particular position xx.

The textbook then takes the limit as Δx→0\Delta x \to 0, turning the sum of rectangles into the exact area under the curve. That limit is the definite integral:

W=∫xixfF(x) dxW = \int_{x_i}^{x_f} F(x) \, dx

Here, WW is the total work done by the variable force as the object moves from xix_i to xfx_f. The symbol ∫\int stands for the integral — the continuous version of the sum. F(x)F(x) is the force at each position, and dxdx is the infinitesimal displacement (the limit of Δx\Delta x). The integral literally means "add up F(x)⋅dxF(x) \cdot dx for every tiny step from xix_i to xfx_f." …

Figure 5.3.b(b) adding all the rectangles, for Δx → 0 the area under the curve equals the work done by F(x).
Fig. 5.3.b — (b) adding all the rectangles, for Δx → 0 the area under the curve equals the work done by F(x).

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure is a simple but powerful visual: a graph with force F(x)F(x) on the vertical axis and position xx on the horizontal axis. The curve drawn is some arbitrary function F(x)F(x) that changes with xx — it is not a straight line, because the force is variable. Two vertical lines are marked at xix_i (the initial position) and xfx_f (the final position). The entire region between the curve, the xx-axis, and these two vertical boundaries is shaded and labelled "Work".

That shaded area is the central idea. For a constant force, work is just F×ΔxF \times \Delta x, which is the area of a rectangle. But when FF changes with xx, you cannot use a single rectangle. The figure shows the textbook's reasoning: if you slice the region under the curve into many thin vertical rectangles of width Δx\Delta x, each rectangle has height approximately F(x)F(x) at that point. The work done over one such slice is roughly F(x) ΔxF(x)\,\Delta x. Summing all these rectangles gives an approximation to the total work. As Δx\Delta x shrinks to zero — the limit shown in the caption — the sum becomes the exact area under the curve.

Important

The physical meaning: the work done by a variable force is the area under the F(x)F(x) vs xx graph between the initial and final positions.

This is not a new formula; it is the definition of work for a one-dimensional variable force. The textbook writes it as:

W=lim⁡Δx→0∑xixfF(x) Δx=∫xixfF(x) dxW = \lim_{\Delta x \to 0} \sum_{x_i}^{x_f} F(x)\,\Delta x = \int_{x_i}^{x_f} F(x)\,dx

Here WW is the work done by the force F(x)F(x) as the object moves from xix_i to xfx_f. The integral symbol ∫\int is just a shorthand for "sum over infinitesimally thin slices". The expression F(x) dxF(x)\,dx represents the work done over an infinitesimal displacement dxdx, and the integral adds up all those infinitesimal contributions.

Note

The figure does not show the rectangles themselves — it shows the final shaded region after the limit has been taken. The caption's mention of "adding all the rectangles, for Δx→0\Delta x \to 0" is the conceptual bridge between the discrete sum and the continuous area. …