Why W = Fd fails for a variable force
For a constant force, work is the simple product \(W = (F\cos\theta)\,d = \vec F\cdot \vec d\): the component of force along the displacement, times the displacement. That definition silently assumes one fixed value of \(F\) acts over the entire path. The moment the force changes with position, the assumption collapses — there is no single \(F\) to multiply by \(d\).
NCERT §5.5 states the situation plainly: "A constant force is rare. It is the variable force, which is more commonly encountered." Stretching a spring, pushing a trunk while tiring, a block entering a rough patch whose resistance depends on position — in each case \(F\) is a function \(F(x)\) of the position \(x\), not a number.
The way out is to recover the constant-force formula on a small enough scale. Over a displacement \(\Delta x\) so short that \(F(x)\) barely changes, the force is approximately constant and the small work done is just the constant-force product on that interval, \(\Delta W = F(x)\,\Delta x\).
From rectangle sum to the integral
Take the displacement from the initial position \(x_i\) to the final position \(x_f\) and chop it into many small intervals of width \(\Delta x\). On each interval treat \(F(x)\) as constant. The small work on one interval is
\(\Delta W = F(x)\,\Delta x\)
Adding the contributions of all the intervals gives the total work as a sum of rectangular areas:
\(W \;\cong\; \displaystyle\sum_{x_i}^{x_f} F(x)\,\Delta x \qquad \text{(NCERT 5.6)}\)
This sum is only approximate, because each rectangle uses one value of \(F\) for the whole strip while the real force drifts across it. Now allow \(\Delta x\) to approach zero. The number of terms grows without limit, but the sum approaches a single definite value — the exact area under the curve. NCERT writes this limit as
\(W \;=\; \displaystyle\lim_{\Delta x \to 0}\sum_{x_i}^{x_f} F(x)\,\Delta x \;=\; \int_{x_i}^{x_f} F(x)\,dx \qquad \text{(NCERT 5.7)}\)
So for a varying force the work done is expressed as a definite integral of force over displacement. The integral sign is nothing more than the limit of the rectangle sum: a continuous tally of \(F\,dx\) slices.
The limiting argument (after NCERT Fig. 5.3). (a) The shaded rectangle of width \(\Delta x\) has area \(F(x)\Delta x = \Delta W\), the work over that small interval; the step profile under-/over-shoots the true curve. (b) As \(\Delta x \to 0\) the rectangle staircase converges to the smooth area under the curve, which equals the work \(W = \int_{x_i}^{x_f} F(x)\,dx\).
Work as the area under the F–x graph
The integral has a picture that is often faster than algebra. Plot the force \(F(x)\) on the vertical axis against displacement \(x\) on the horizontal axis. Each \(F\,dx\) slice is the area of a vertical sliver under the curve, so the whole integral is the area enclosed between the curve and the x-axis, from \(x_i\) to \(x_f\). NIOS §6.2 states it in one line: "Work is numerically equal to the area under the F versus x graph."
This is powerful for NEET, because many problems hand you a graph of straight-line segments rather than an equation. You then do not integrate at all — you compute the area of rectangles, triangles and trapeziums by school geometry and add them up.
| Shape on the F–x graph | Force behaviour | Work = area |
|---|---|---|
| Rectangle | Constant force over the interval | \(W = F \times \Delta x\) |
| Triangle (from zero) | Force rising/falling linearly from \(0\) | \(W = \tfrac12 \times \text{base} \times \text{height}\) |
| Trapezium | Force changing linearly between two non-zero values | \(W = \tfrac12 (F_1 + F_2)\,\Delta x\) |
| Area below the x-axis | Force negative (opposes displacement) | Negative work |
The sign of the area
Area is not blindly positive. Where the force is plotted above the x-axis (\(F > 0\), acting along the displacement), the area counts as positive work. Where the force is plotted below the axis (\(F < 0\), opposing the displacement), the area counts as negative work. NCERT puts it directly in Example 5.5: "The area on the negative side of the force axis has a negative sign."
This is the same sign rule you already know from the constant-force case — \(\cos\theta\) is positive for \(0^\circ \le \theta < 90^\circ\) and negative for \(90^\circ < \theta \le 180^\circ\) — now read off a graph. A friction force, which acts opposite to motion, sits below the axis and contributes negative work.
If the sign convention for \(\cos\theta\) feels shaky, revisit work — definition and calculation before pushing on.
Worked example 1 — reading work off an F–x graph
A woman pushes a trunk on a rough railway platform. She applies a force of 100 N over the first 10 m. Then she tires, and her applied force falls linearly with distance to 50 N at \(x = 20\) m. The frictional force, of constant magnitude 50 N, opposes the motion throughout. Find the work done by the applied force and by friction over the full 20 m.
Set up the graph. The applied force \(F\) sits above the x-axis: a flat segment at 100 N from \(x = 0\) to \(x = 10\) m (a rectangle), then a straight drop from 100 N to 50 N from \(x = 10\) to \(x = 20\) m (a trapezium). Friction \(f\) is constant at 50 N but opposes motion, so it is drawn below the axis as a rectangle of height \(-50\) N spanning the full 20 m.
Work by the woman = area of the rectangle + area of the trapezium:
\(W_F = (100 \times 10) + \tfrac12 (100 + 50)\times 10 = 1000 + 750 = 1750~\text{J}\)
Work by friction = area of the rectangle below the axis:
\(W_f = (-50)\times 20 = -1000~\text{J}\)
Answer: \(W_F = +1750~\text{J}\), \(W_f = -1000~\text{J}\). No integration was needed — only the areas of a rectangle, a trapezium and a (negative) rectangle.
F–x plot for Example 5.5 (after NCERT Fig. 5.4). The applied force gives a rectangle (1000 J) plus a trapezium (750 J) above the axis, totalling \(W_F = 1750\) J. The constant 50 N friction, opposing motion, is a rectangle below the axis with \(W_f = -1000\) J.
The spring force as a variable force
The cleanest variable force in the NEET syllabus is the spring force. A block on a smooth surface is attached to a light spring fixed to a wall. For an ideal spring the force is proportional to the displacement \(x\) from the equilibrium position and points back toward it. This is Hooke's law:
\(F_s = -kx\)
The constant \(k\) is the spring constant, with unit \(\text{N m}^{-1}\); a stiff spring has large \(k\), a soft spring small \(k\). The minus sign encodes the restoring direction: for a stretched spring \(x > 0\) and \(F_s < 0\); for a compressed spring \(x < 0\) and \(F_s > 0\). Because \(F_s\) changes with \(x\), the work it does must be found by integration — or by the area of the triangle on its \(F_s\)–\(x\) plot.
Pulling the block out to an extension \(x_m\), the work done by the spring force is
\(W_s = \displaystyle\int_0^{x_m} F_s\,dx = -\int_0^{x_m} kx\,dx = -\tfrac12 k x_m^{2} \qquad \text{(NCERT 5.15)}\)
The result is negative: while you stretch the spring, the spring force opposes the displacement, so it does negative work. The external pulling force, overcoming the spring, does the opposite, \(+\tfrac12 k x_m^{2}\) (NCERT 5.16). That positive work is stored as elastic potential energy. NIOS §6.2.1 reaches the same \(\tfrac12 k x_m^{2}\) by a neat shortcut: since the force rises linearly from 0 to \(kx_m\), its average over the stretch is \(\tfrac12 kx_m\), and average force times displacement gives \(\tfrac12 kx_m \cdot x_m = \tfrac12 k x_m^{2}\).
\(F_s\)–\(x\) plot for an ideal spring (after NCERT Fig. 5.7d). Hooke's law \(F_s = -kx\) is a straight line through the origin. Stretching to \(x_m\) traces the triangle below the axis; its area, \(\tfrac12 x_m \cdot k x_m\), carries a negative sign because force and displacement have opposite signs — hence \(W_s = -\tfrac12 k x_m^{2}\).
Two further facts follow straight from the integral, and both are heavily examined. First, for a move from any \(x_i\) to any \(x_f\),
\(W_s = -\displaystyle\int_{x_i}^{x_f} kx\,dx = \tfrac12 k x_i^{2} - \tfrac12 k x_f^{2} \qquad \text{(NCERT 5.17)}\)
which depends only on the end points. Second, over a cyclic process that returns to \(x_i\), the work is zero (NCERT 5.18). These two properties are the definition of a conservative force, and the stored energy is the elastic potential energy \(V(x) = \tfrac12 k x^{2}\) (NCERT 5.19).
Worked example 2 — work to stretch a spring
A mass of 2 kg is attached to a light spring of force constant \(k = 100~\text{N m}^{-1}\). Calculate the work done by an external force in stretching the spring by 10 cm.
Use the spring-work integral / triangle area. The external force overcomes the spring, doing positive work equal in magnitude to \(\tfrac12 k x^{2}\), with \(x = 10~\text{cm} = 0.1~\text{m}\):
\(W = \tfrac12 k x^{2} = \tfrac12 \times 100 \times (0.1)^{2} = 50 \times 0.01 = 0.5~\text{J}\)
Answer: the external force does \(+0.5~\text{J}\). By the opposite-sign rule, the spring's restoring force does \(-0.5~\text{J}\) over the same stretch. The 2 kg mass is irrelevant here — the work to stretch a spring depends only on \(k\) and the extension, not on what is attached.
Worked example 3 — an inverse force law
A block of mass \(m = 1\) kg moves on a horizontal surface with speed \(v_i = 2~\text{m s}^{-1}\) and enters a rough patch from \(x = 0.10\) m to \(x = 2.01\) m. The retarding force in this range is \(F_r = -\dfrac{k}{x}\) with \(k = 0.5~\text{J}\), and zero outside it. Find the final kinetic energy and speed as the block crosses the patch.
This force is not a straight line — its graph is a hyperbola, so the area cannot be read as a triangle. Integrate directly. By the work–energy theorem for a variable force, \(K_f - K_i = \int F_r\,dx\):
\(K_f = \tfrac12 m v_i^{2} + \displaystyle\int_{0.1}^{2.01}\!\left(-\frac{k}{x}\right)dx = \tfrac12 m v_i^{2} - k\,\ln\!\frac{2.01}{0.1}\)
With \(\tfrac12 m v_i^{2} = \tfrac12 (1)(2)^2 = 2~\text{J}\) and \(\ln(20.1) \approx 3\):
\(K_f = 2 - 0.5 \times \ln(20.1) = 2 - 1.5 = 0.5~\text{J}\)
Final speed from \(K_f = \tfrac12 m v_f^{2}\): \(v_f = \sqrt{2K_f/m} = \sqrt{2 \times 0.5 / 1} = 1~\text{m s}^{-1}\).
Answer: \(K_f = 0.5~\text{J}\), \(v_f = 1~\text{m s}^{-1}\). For a curved force law the integral is unavoidable; here it produces a natural logarithm. (Note \(\ln X = 2.303\,\log_{10}X\), the logarithm to base \(e\).)
Examples 1 to 3 cover the full range of NEET demands on this subtopic: a piecewise-linear graph read as areas, a linear spring law integrated to \(\tfrac12 k x^2\), and a genuinely curved law that forces a direct integration. The same definition, \(W = \int F\,dx\), drives all three.
Typical errors
- Using \(W = Fd\) with a varying force. Valid only when \(F\) is constant. Otherwise integrate or take the area.
- Forgetting the sign of the area. Force below the x-axis means negative work. Friction and the spring's restoring force routinely sit below the axis.
- Confusing spring-force work with external-force work. Stretching: spring does \(-\tfrac12 k x_m^2\), the external agent does \(+\tfrac12 k x_m^2\). They are equal and opposite.
- Treating spring energy as linear in \(x\). It is \(\tfrac12 k x^2\); quadrupling the extension multiplies the energy by sixteen.
- Reading the net area when one force is asked. Take only the strip of the named force.
- Forcing a triangle onto a curved law. A hyperbolic or other non-linear \(F(x)\) needs the actual integral, not a half-base-height shortcut.