Work–energy theorem for a variable force

Push a shopping trolley across a flat floor with a steady shove and the work-energy bookkeeping is a one-line multiplication: force times distance equals the gain in kinetic energy. But stretch a bow, compress a spring-loaded toy gun, or launch a rocket that loses mass and feels changing air drag, and the force refuses to sit still — it grows, shrinks, or reverses as the object moves. Does the tidy rule $W_{net} = \Delta K$ survive when the force itself is a moving target? It does — and proving that it does, cleanly, using nothing beyond the calculus you already have from the variable-force work topic, is the entire point of this article. The trick is a single, elegant substitution: instead of chasing force as a function of time, Newton's second law is rewritten so that velocity is treated as a function of position. That one move turns a hard problem — force changing continuously along a path — into an ordinary integral that any Class 11 student can evaluate, and it lands, remarkably, on exactly the same equation you already trust for constant forces. This matters far beyond the exam hall. Every spring, every bow, every rubber band, and even the thrust of a rocket burning fuel all deliver variable forces. If the work-energy theorem broke down every time a force changed with position, it would be almost useless — physics would need a different tool for every different kind of push. Instead, one theorem, proved once and for all in this article, quietly handles constant forces, spring forces, drag that grows with speed, and any force you can write down as a function of $x$ . You will derive it from scratch, apply it to a spring-launched block, and then push the idea further with a two-force competitive-level problem that mixes a variable spring force with constant friction — exactly the kind of combination that shows up in JEE and NEET mechanics.

Key takeaways

The work-energy theorem $W_{net} = \Delta K$ is not limited to constant forces — it holds for any force that depends on position, $F(x)$ .
The proof rewrites Newton's second law using $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ (the chain rule), turning $W = \int F\,dx$ into $W = \int mv\,dv$ .
Evaluating that integral from $v_i$ to $v_f$ gives $W = \tfrac12 mv_f^2 - \tfrac12 mv_i^2 = \Delta K$ — identical in form to the constant-force result.
To use the theorem for a variable force, first find $W = \int F(x)\,dx$ (or the area under the $F$ – $x$ graph) for every force acting, then set the total equal to $\Delta K$ .
Spring problems are the classic application: $W_{\text{spring}} = \tfrac12 k x^2$ feeds directly into $\Delta K$ to give launch speeds.
When more than one force acts — say a spring and friction — add the works algebraically (with correct signs) before equating the sum to $\Delta K$ ; this is where most marks are lost.
Key idea: because $F = mv\,dv/dx$ , the work integral $\int F\,dx$ collapses to $\int mv\,dv = \Delta K$ for any position-dependent force, so $W_{net} = \Delta K$ is a universal statement, not a constant-force special case.

The core insight

Substituting $a = v\,dv/dx$ into Newton's second law turns the work integral $\int F(x)\,dx$ into $\int mv\,dv$ , proving that $W_{net} = \Delta K$ holds for any force that varies with position, not just constant ones.

Where it fits in the syllabus

This article sits inside the CBSE Class 11 unit Work, Energy and Power, directly after the topic of computing work done by a variable force via $W = \int F\,dx$ and after the basic (constant-force) work-energy theorem. It generalises that constant-force result to any position-dependent force using only elementary integration — no vector calculus, no vector cross products, nothing beyond the algebra and single-variable integration already used for variable-force work. The syllabus expects you to follow the derivation, recognise it as a genuine extension (not a different theorem), and apply it to problems such as spring launches — including problems, like the competitive example below, that combine a variable force with a constant one such as friction.

The physical picture

Newton's second law connects force to acceleration at every instant: $F = ma$ . When $F$ is constant, $a$ is constant too, and the constant-acceleration kinematics equations do all the work. But once $F$ varies with position — a spring pushing harder the more it is compressed, for instance — acceleration also varies with position, and $v = u + at$ stops being valid. The way out is a change of variable: instead of treating velocity as a function of time, treat it as a function of position, and let the chain rule connect the two.

a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\,\frac{dv}{dx}

where: $a$ is instantaneous acceleration, $v$ is instantaneous speed, $x$ is position along the line of motion, and $dv/dx$ is how quickly speed changes with position (not with time). Physically, this equation says nothing new about the motion — it is the same acceleration, just re-expressed in terms of how fast velocity is changing per metre travelled rather than per second elapsed. That re-expression is the whole trick: it lets force, which is naturally a function of position for springs and similar systems, sit next to a $dv$ term that can be integrated directly. Substituting into $F = ma$ gives $F = mv\,dv/dx$ , and multiplying both sides by $dx$ turns the force-versus-position relationship into something you can integrate term by term as the particle moves from $x_i$ to $x_f$ , with matching speeds $v_i$ and $v_f$ :

W = \int_{x_i}^{x_f} F\,dx = \int_{x_i}^{x_f} mv\,\frac{dv}{dx}\,dx = \int_{v_i}^{v_f} mv\,dv

where: the left-hand integral is the familiar area-under-the- $F$ – $x$ -graph definition of work; the right-hand integral is now purely in terms of speed, with the position variable having cancelled out entirely. This is the pivotal step — a hard geometric problem (area under a curved, changing force graph) has been converted into an easy algebraic one (integrating $v\,dv$ ).

Figure

A force-position graph showing a curved line for a variable force above the axis and the shaded area under the curve from the starting position to the final position, with a small inset arrow showing the same shaded area re-expressed as a triangle on a velocity axis representing the kinetic energy change

Carrying out the right-hand integral term by term:

W = m\int_{v_i}^{v_f} v\,dv = m\left[\frac{v^2}{2}\right]_{v_i}^{v_f} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = K_f - K_i

where: $K_i = \tfrac12 mv_i^2$ and $K_f = \tfrac12 mv_f^2$ are the initial and final kinetic energies. Since the left-hand side, $W$ , was defined as the work done by the net force, the result is exactly:

W_{net} = \Delta K

This is the same equation used for a constant force — the theorem's form never changed. What changed is only the recipe for computing $W$ : for a constant force it was $Fd\cos\theta$ ; for a variable force it is $\int F\,dx$ (an area, not a simple product). The physical content of the theorem — that the total work done on a particle by every force acting on it equals the change in its kinetic energy — is completely general, and it makes no reference at all to whether the force is constant or variable. That generality is exactly why the theorem is one of the most useful tools in mechanics: a single equation covers gravity, springs, friction, and any other force you can express as a function of position.

Derivation AProving the work-energy theorem for a force that varies with position

The goal of this derivation is to show, starting only from Newton's second law and the definition of work as an integral, that the relation $W_{net} = \Delta K$ holds exactly for a particle moving under any force $F(x)$ that depends on position alone — not merely for the special case of a constant force — using nothing beyond the chain rule and a single elementary integral, so that the same theorem already trusted for constant-force problems can be applied without modification to springs, variable drag, and any other position-dependent push or pull.

Step 1. Start from Newton's second law for a particle of mass $m$ moving along the x-axis under a net force $F(x)$ that may depend on position: $F = ma$ , where $a = dv/dt$ is the instantaneous acceleration.

Step 2. Rewrite the acceleration using the chain rule so that it is expressed in terms of position instead of time: $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ , since $dx/dt$ is simply the velocity $v$ . This substitution is valid at every instant of the motion and introduces no approximation.

Step 3. Substitute this expression for $a$ into Newton's second law to get $F = mv\dfrac{dv}{dx}$ , a relation that ties the force directly to how velocity changes with position — exactly the form needed to combine with the definition of work as $\int F\,dx$ .

Step 4. Write the work done by this force as the particle moves from position $x_i$ (with speed $v_i$ ) to position $x_f$ (with speed $v_f$ ): $W = \displaystyle\int_{x_i}^{x_f} F\,dx = \int_{x_i}^{x_f} mv\,\frac{dv}{dx}\,dx$ .

Step 5. Cancel the $dx$ against the $dx$ inside $dv/dx$ , changing the variable of integration from position to speed: $W = \displaystyle\int_{v_i}^{v_f} mv\,dv$ . Position has disappeared from the integral entirely — the work now depends only on the initial and final speeds, whatever twisting path $F(x)$ took to produce them.

Step 6. Evaluate this elementary integral: $W = m\left[\dfrac{v^2}{2}\right]_{v_i}^{v_f} = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$ .

Step 7. Recognise the two terms as kinetic energies, $K_i = \tfrac12 mv_i^2$ and $K_f = \tfrac12 mv_f^2$ , so that $W = K_f - K_i = \Delta K$ . Since no assumption was made anywhere about $F$ being constant — only that it could be written as some function $F(x)$ — this result, $W_{net} = \Delta K$ , holds for any position-dependent force, completing the generalisation of the work-energy theorem.

Worked examples

Board level - variable force $F(x) = (6-2x)\ \text{N}$

A particle of mass $2\ \text{kg}$ moves along the x-axis under a force $F(x) = (6-2x)\ \text{N}$ , with $x$ measured in metres. It starts at $x=0$ with speed $3\ \text{m/s}$ . Find its speed at $x = 3\ \text{m}$ .

Step 1. Compute the work done using the integral definition: $W = \displaystyle\int_0^3 (6-2x)\,dx = \left[6x - x^2\right]_0^3 = (18-9)-0 = 9\ \text{J}$ .

Step 2. Apply the (variable-force) work-energy theorem just derived: $W = K_f - K_i$ .

Step 3. Compute the initial kinetic energy: $K_i = \dfrac12(2)(3)^2 = 9\ \text{J}$ .

Step 4. Solve for the final kinetic energy, then the final speed: $K_f = W + K_i = 9+9 = 18\ \text{J}$ , so $\dfrac12(2)v_f^2 = 18 \Rightarrow v_f^2 = 18 \Rightarrow$ $v_f \approx 4.24\ \text{m/s}$ .

Competitive level - spring launch followed by a frictional floor

A spring with force constant $k = 400\ \text{N/m}$ is compressed by $0.1\ \text{m}$ and released, pushing a $0.5\ \text{kg}$ block that starts at rest against it. The spring section of the floor is frictionless, but beyond the spring's natural length the floor has friction with coefficient $\mu = 0.2$ . Find (a) the block's speed as it leaves the spring, and (b) how far it slides before stopping.

Step 1. Find the work done by the spring as it expands from compression $x=0.1\ \text{m}$ to its natural length, using $W_{\text{spring}} = \tfrac12 kx^2$ (the area under the spring's $F$ – $x$ graph): $W_{\text{spring}} = \dfrac12(400)(0.1)^2 = 2\ \text{J}$ .

Step 2. Apply the work-energy theorem for this stage, with $K_i = 0$ (block starts at rest) and no friction yet acting: $W_{\text{spring}} = K_f - 0 = 2\ \text{J}$ .

Step 3. Solve for the launch speed: $\dfrac12(0.5)v_f^2 = 2 \Rightarrow v_f^2 = 8 \Rightarrow$ $v_f \approx 2.83\ \text{m/s}$ , answering part (a).

Step 4. For part (b), treat the frictional stretch as a second stage where the only force doing work is friction, which is constant in magnitude but always opposes motion: $f = \mu mg = (0.2)(0.5)(9.8) = 0.98\ \text{N}$ , so $W_{\text{friction}} = -f\,d$ for a sliding distance $d$ .

Step 5. Apply the work-energy theorem to this second stage, starting from the kinetic energy found in Step 3 and ending at rest: $W_{\text{friction}} = K_{\text{final}} - K_{\text{initial}} = 0 - \dfrac12(0.5)(8) = -2\ \text{J}$ .

Step 6. Solve for the sliding distance: $-0.98\,d = -2 \Rightarrow$ $d \approx 2.04\ \text{m}$ , answering part (b). Notice how the same theorem handled the variable spring force in Steps 1-3 and the constant friction force in Steps 4-6, simply by recomputing $W$ with the method appropriate to each force and applying $W=\Delta K$ separately at each stage. Timing note: budget about 3-4 minutes for the launch-speed part alone in a board exam; a full two-stage problem like this, worth 4-5 marks in JEE Main-style tests, should take no more than 5-6 minutes if you keep the two stages clearly separated on paper rather than trying to merge them into one energy equation.

Common traps

Believing the work-energy theorem is a "constant-force-only" formula — the derivation above proves it explicitly for any $F(x)$ ; only the method of computing $W$ changes, never the theorem itself.
Reaching for $v = u+at$ or other constant-acceleration kinematics when the force (and hence acceleration) varies with position — those equations are simply invalid here; only the work-energy theorem or a fresh integration of Newton's law will work cleanly.
Sign errors in $F(x)$ before integrating, especially forgetting the minus sign in a restoring force like $F=-kx$ — this silently flips the sign of the computed work.
Treating a multi-force, multi-stage problem (like the competitive example) as a single energy equation instead of tracking each stage's net work separately — mixing a spring stage and a friction stage into one line is the single most common way marks are lost on this topic.
Forgetting that "net" work means the sum of the work done by every force acting, not just the one force that happens to be varying — a constant force like gravity or friction alongside a variable one must still be added in.

You should be able to…

Derive $W_{net} = \Delta K$ for a position-dependent force starting from $F=ma$ and the chain-rule substitution $a = v\,dv/dx$ .
Explain why the integral $\int F\,dx$ collapses to $\int mv\,dv$ and why position drops out of the final result.
Compute the work done by a stated variable force using $W = \int F\,dx$ , and apply the theorem to find an unknown speed or position.
Recognise that $W_{net} = \Delta K$ has exactly the same form for constant and variable forces — only the recipe for $W$ differs.
Handle multi-stage problems where a variable force (like a spring) and a constant force (like friction) act in sequence, applying the theorem stage by stage.
Spot and avoid the classic sign error that comes from mishandling a restoring force such as $F=-kx$ .

Ask the tutor: Want to see the chain-rule substitution $a = v\,dv/dx$ worked through slowly with a diagram, or try a fresh variable-force problem of your own to check you can apply the three-step recipe?

If the theorem works for any force, why do textbooks even bother deriving the constant-force version separately?

Mostly for pedagogical order — the constant-force proof uses only algebra ( $v^2 = u^2+2as$ combined with $F=ma$ ), which is introduced before integration, so it is taught first as a warm-up. Once you have integration, the variable-force derivation in this article supersedes it: setting $F(x)$ equal to a constant in the general proof reproduces the earlier constant-force result exactly. There are not two theorems, only one theorem proved at two different levels of mathematical tool.

Does this theorem tell me the direction of motion or just the speed?

Just the speed (or, equivalently, the magnitude of kinetic energy change) — kinetic energy is $\tfrac12 mv^2$ , which depends on $v^2$ and therefore cannot distinguish a particle moving in the $+x$ direction from one moving in the $-x$ direction with the same speed. If a question asks for velocity as a vector (direction included), you need extra information from the problem, such as which way the particle was pushed, not the work-energy theorem alone.

Can I use this theorem when friction is also acting, since friction is not really "position-dependent" in the same way as a spring?

Yes — friction's magnitude is usually constant while it acts (as in the competitive example above), which makes it a special, simple case of a force that could still be written as a function of $x$ (a constant function). The general theorem $W_{net}=\Delta K$ never required $F(x)$ to actually vary; it only required that a work integral could be written down. Add the work done by friction ( $-f d$ , since friction always opposes displacement) to the work done by any other forces present, and set that total equal to $\Delta K$ .

Is there a shortcut so I don't have to redo the full derivation every time in an exam?

For applying the theorem to a problem, no — you just quote the two-step recipe: compute $W=\int F\,dx$ (or add up the areas under each force's $F$ – $x$ graph) for every force acting, then set the sum equal to $\Delta K = \tfrac12 mv_f^2 - \tfrac12 mv_i^2$ . The full chain-rule derivation shown above is worth memorising step-by-step only because CBSE board papers sometimes ask you to "derive the work-energy theorem for a variable force" as a standalone 3-5 mark question, separate from any numerical problem.

Work–energy theorem for a variable force

Where it fits in the syllabus

The physical picture

Worked examples

Board level - variable force $F(x) = (6-2x)\ \text{N}$

Competitive level - spring launch followed by a frictional floor

Common traps

You should be able to…

Before this

Where this leads

Work–energy theorem for a variable force

Where it fits in the syllabus

The physical picture

Worked examples

Board level - variable force F(x)=(6−2x) NF(x) = (6-2x)\ \text{N}F(x)=(6−2x) N

Competitive level - spring launch followed by a frictional floor

Common traps

You should be able to…

Before this

Where this leads

Board level - variable force $F(x) = (6-2x)\ \text{N}$