Work-energy theorem for a variable force

What does the theorem say for a variable force?

In plain words: the work-energy theorem, that the work done by the net force equals the change in kinetic energy, is not limited to constant forces. For a constant-mass particle in an inertial frame it holds for forces varying with position, time or velocity, provided their work is evaluated along the actual motion. This lesson treats one-dimensional position-dependent forces. The proof is a short piece of calculus, and it shows that the theorem is Newton’s second law “added up” along the path.

Here K_i and K_f are the kinetic energies at the positions x_i and x_f, and the integral is the work done by the net force F(x), the area under its force–displacement graph (see work done by a variable force).

Proof using calculus

Take a body of constant mass m moving along the x-axis under a net force F that may change with position. Its kinetic energy is $K = \tfrac{1}{2}mv^{2}$.

Step 1: rate of change of kinetic energy. Differentiate with respect to time, using the chain rule on v²:

Step 2: use Newton’s second law. Since m dv/dt = F,

Step 3: cancel the time. Multiply both sides by dt:

The small change in kinetic energy over a small displacement dx is the small work done, F dx. Writing dK as mv dv gives mv dv = F dx. Where velocity can be treated as a differentiable function of position, this yields v dv/dx = a. At a turning point, dv/dx may be undefined; the time-based derivation above remains valid.

Step 4: integrate from the initial position to the final position:

where W is the work done by the variable net force. This proves the theorem for a force that varies with position (NCERT gives the proof in one dimension).

What the theorem leaves out

The theorem is powerful because it skips straight from force to speed. But it pays for this in two ways:

• No time. Newton’s second law links force and acceleration at every instant. The theorem is an integral over the whole motion, so it says nothing directly about how long the motion takes.
• No direction. In two or three dimensions, Newton’s second law is a vector equation. The work-energy theorem is a scalar equation, so it cannot tell you the direction of motion, only the speed.

So use the theorem for “how fast” and “how far” questions, and Newton’s second law for “how long” and “in which direction” questions.

Where it helps

• Rough patches. When friction or resistance depends on position (a rough patch, a muddy stretch), integrate it to find the work, then the final speed.
• Springs and elastic cords. The force grows with the stretch, but the work, and so the change in speed, comes straight from the integral.
• Stopping distance against a growing force. A body sliding into a region where the retarding force grows with distance stops where its kinetic energy has all been used up.
• Gravity over large heights. Gravity weakens with distance from the Earth’s centre, so the work it does on a body moving straight up or down over a large height is an integral, and the theorem then gives the change in the body’s kinetic energy.

Solved examples

Common mistakes to avoid

Quick revision summary

• The work-energy theorem holds for forces that vary with position: the change in kinetic energy is the integral of the net force over the displacement.
• Proof: the rate of change of kinetic energy is mv dv/dt = Fv = F dx/dt, so dK = F dx; integrate.
• The theorem is an integral form of Newton’s second law; it gives speeds but not time or direction.
• Use it for “how fast” and “how far” questions when the force depends on position.

Questions students ask

Why is the work-energy theorem called an integral form of Newton’s second law?

The work-energy theorem is called an integral form of Newton’s second law because it is the second law, m dv/dt = F, integrated over the displacement: with dv/dt = v dv/dx it becomes mv dv = F dx, which adds up to $K_f - K_i = \int F\,dx$. The integration keeps the link between force and speed but loses the time and, in more than one dimension, the direction. So it tells you how fast a body moves at each position, not when it gets there.

Is the work-energy theorem valid in a non-inertial frame?

The work-energy theorem holds in a non-inertial frame only if you include the work done by the pseudo force. Newton’s second law, from which the theorem comes, holds in a translating frame with non-rotating axes and acceleration a₀ after a pseudo force −m a₀ is added to the real forces. Rotating frames require the additional inertial-force terms as well. With all applicable pseudo-force work counted in the net work, and all speeds and displacements measured in that frame, the change in kinetic energy equals the net work.

Does the work-energy theorem hold in two and three dimensions?

Yes, the work-energy theorem holds in two and three dimensions. In vector form dK = F · dr, where F is the net force, so, integrating along the actual path,

The path is essential for a general force. Only the component of a force along the motion does work: for an ideal pendulum with a fixed pivot and an inextensible string, tension is perpendicular to the bob’s velocity and does no work. If air resistance and other forces are absent, gravity alone changes the speed.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A 4 kg body starts from rest at x = 1 m, and a net force F = 6x N acts on it along the x-axis. Find its speed at x = 3 m.

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   $$\begin{aligned} W &= \int_1^{3} 6x\,dx \\ &= \Big[3x^{2}\Big]_1^{3} \\ &= 3(9-1) = 24\ \text{J} \end{aligned}$$

   So $\tfrac{1}{2}(4)v^{2}$ is 24 J, giving v² = 12 and v ≈ 3.46 m s^⁻¹. Starting at x = 0 with zero velocity would instead leave the body at rest, since its initial force would also be zero.

2. Q2. State two things that Newton’s second law tells you about a motion but the work-energy theorem does not.

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   The time taken (the theorem is integrated over the motion, so the time is not directly available), and the direction of motion in two or three dimensions (the theorem is a scalar equation, while the second law is a vector equation).

3. Q3. A 2 kg block moving at 5 m s^⁻¹ enters a rough region where the friction force grows as F = −3x N. How far does it go before stopping?

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   Its kinetic energy is $\tfrac{1}{2}(2)(5)^{2}$, that is 25 J. The work done after a distance x is −1.5x², so it stops when 1.5x² = 25:

   $$x = \sqrt{\frac{25}{1.5}} \approx 4.08\ \text{m}$$