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Work and Variable Forces · Tutorial 470 of 998

Deriving the Work-Energy Theorem

See why net work equals the change in kinetic energy, and use that relationship to connect force, position, mass, and speed.

Intermediate 10 min read

What You'll Learn

  • Define net work along a one-dimensional motion.
  • Derive the work-energy theorem from Newton’s second law.
  • Relate a position-dependent net force to work over a position interval.
  • Use net work to calculate a change in kinetic energy and determine speed.
  • Interpret positive, negative, and piecewise work with correct signs.

From Force to Change in Motion

A force can change an object's motion, but the effect of a force depends on how far the object moves while the force acts. Work measures this interaction between force and displacement. The work-energy theorem makes the connection precise: the total work done by the net force equals the change in the object's kinetic energy.

This relationship follows directly from Newton's second law. In one dimension, it can be derived using the chain rule, without requiring the force to be constant. It applies to variable forces as well, provided the motion and force satisfy the assumptions stated below.

Work and Kinetic Energy in One Dimension

Let a particle of constant mass \(m\) move along the \(x\)-axis. Its position is \(x(t)\), its velocity is \(v(t)=x'(t)\), and its acceleration is \(a(t)=v'(t)\). The net force \(F_{\text{net}}(t)\) is the signed sum of all forces along the axis. A positive force points in the positive \(x\)-direction; a negative force points in the opposite direction.

Definition: The work done by a one-dimensional force \(F\) during motion from time \(t_i\) to time \(t_f\) is \(W=\int_{t_i}^{t_f} F(t)v(t)\,dt\). Since \(dx=v(t)\,dt\), this is also written \(W=\int F\,dx\) along the motion. Net work is the work done by the net force, or equivalently the sum of the works done by all the forces.

Displacement and force carry signs in this definition. A force in the direction of motion contributes positive work; a force opposite the direction of motion contributes negative work. If the object reverses direction, the signed displacement changes direction too, and the integral accounts for that.

The kinetic energy of a particle of mass \(m\) and speed \(|v|\) is \(K=\frac12 mv^2\). Because velocity is squared, kinetic energy is nonnegative and does not encode the direction of travel. The work-energy theorem concerns the change in kinetic energy, not kinetic energy by itself.

$$ \Delta K=K_f-K_i=\frac12 mv_f^2-\frac12 mv_i^2. $$

Deriving the Work-Energy Theorem

Assume the particle has constant mass, its position is twice differentiable over the time interval considered, and Newton's second law applies in an inertial frame. The force may vary with time or position. The key step is to differentiate kinetic energy with respect to time and use Newton's second law.

1
Differentiate kinetic energy.
Since \(K(t)=\frac12 mv(t)^2\) and \(m\) is constant, the chain rule gives \(\frac{dK}{dt}=mv(t)\frac{dv}{dt}\).
2
Apply Newton's second law.
Because \(\frac{dv}{dt}=a(t)\) and \(F_{\text{net}}(t)=ma(t)\), it follows that \(\frac{dK}{dt}=F_{\text{net}}(t)v(t)\).
3
Integrate over the motion.
Integrating from \(t_i\) to \(t_f\) gives \(K_f-K_i=\int_{t_i}^{t_f}F_{\text{net}}(t)v(t)\,dt\). By the definition of work, the integral is the net work.
$$ \begin{aligned} \frac{dK}{dt} &=\frac{d}{dt}\left(\frac12 mv(t)^2\right)\\ &=mv(t)\frac{dv}{dt}\\ &=ma(t)v(t)\\ &=F_{\text{net}}(t)v(t). \end{aligned} $$
$$ \begin{aligned} K_f-K_i &=\int_{t_i}^{t_f}\frac{dK}{dt}\,dt\\ &=\int_{t_i}^{t_f}F_{\text{net}}(t)v(t)\,dt\\ &=W_{\text{net}}. \end{aligned} $$
Theorem: For a particle of constant mass moving in one dimension under Newton's second law, the net work done on the particle over a time interval equals its change in kinetic energy: \(W_{\text{net}}=K_f-K_i=\frac12 mv_f^2-\frac12 mv_i^2\).

This derivation does not divide by velocity, so it remains valid if the particle is momentarily at rest or changes direction during the interval. If the net force depends only on position, \(F_{\text{net}}=F_{\text{net}}(x)\), the work can also be calculated from the position integral \(\int_{x_i}^{x_f}F_{\text{net}}(x)\,dx\). In one dimension, this integral gives the same work even if the path reverses, because a position-dependent force has an antiderivative on the interval. If the force depends on time as well as position, use the integral along the actual motion, \(\int F_{\text{net}}(t)\,dx\).

Using the Theorem to Find Speed

When the net work and initial speed are known, the theorem gives the final kinetic energy. Rearranging the equation gives the final speed, provided the expression under the square root is nonnegative:

$$ v_f^2=v_i^2+\frac{2W_{\text{net}}}{m}, \qquad |v_f|=\sqrt{v_i^2+\frac{2W_{\text{net}}}{m}}. $$

This determines speed, not direction. To determine the sign of the final velocity, use information about the direction of motion and the forces along the interval. If the right-hand side is negative, the stated motion through the entire interval is not physically possible with those initial conditions and that net work.

Worked Example: Constant Net Force

A \(3\,\text{kg}\) particle moves from \(x_i=2\,\text{m}\) to \(x_f=7\,\text{m}\) under a constant net force of \(6\,\text{N}\) in the positive direction. Its initial velocity is \(2\,\text{m/s}\). Find the net work and final speed.

The displacement is \(7-2=5\,\text{m}\), so the work is

$$ W_{\text{net}}=(6\,\text{N})(5\,\text{m})=30\,\text{J}. $$

The initial kinetic energy is

$$ K_i=\frac12(3\,\text{kg})(2\,\text{m/s})^2 =\frac12(3)(4)\,\text{J}=6\,\text{J}. $$

Thus \(K_f=K_i+W_{\text{net}}=6+30=36\,\text{J}\). Using \(K_f=\frac12 mv_f^2\),

$$ 36\,\text{J}=\frac12(3\,\text{kg})v_f^2, \qquad v_f^2=24\,\text{m}^2/\text{s}^2, \qquad |v_f|=2\sqrt{6}\,\text{m/s}\approx4.90\,\text{m/s}. $$

The force points in the positive direction throughout, and the particle starts moving in that direction, so its final velocity is positive. Check: \(\frac12(3)(2\sqrt6)^2=36\,\text{J}\), and \(36-6=30\,\text{J}\), matching the calculated work.

Variable and Piecewise Net Forces

For a position-dependent force, the work is found by integrating the force over position. The limits must follow the direction of displacement, and the force must be the net force if the theorem is to give the total change in kinetic energy. A force that opposes motion may contribute negative work, reducing kinetic energy; it does not necessarily make the total work negative.

Worked Example: A Position-Dependent Net Force

A \(2\,\text{kg}\) particle moves in the positive direction from \(x=0\,\text{m}\) to \(x=3\,\text{m}\). Its net force is \(F_{\text{net}}(x)=(4\,\text{N/m})x\), and its initial speed is \(1\,\text{m/s}\). Find the work and final speed.

Integrate the force over the position interval:

$$ \begin{aligned} W_{\text{net}} &=\int_0^3 (4\,\text{N/m})x\,dx\\ &=(4\,\text{N/m})\left[\frac{x^2}{2}\right]_0^3\\ &=(4\,\text{N/m})\left(\frac{9-0}{2}\,\text{m}^2\right) =18\,\text{J}. \end{aligned} $$

The initial kinetic energy is \(\frac12(2\,\text{kg})(1\,\text{m/s})^2=1\,\text{J}\). Therefore \(K_f=1+18=19\,\text{J}\), and

$$ 19\,\text{J}=\frac12(2\,\text{kg})v_f^2, \qquad |v_f|=\sqrt{19}\,\text{m/s}\approx4.36\,\text{m/s}. $$

The force is positive for \(x>0\), so the particle continues in the positive direction and \(v_f=\sqrt{19}\,\text{m/s}\). Check: its final kinetic energy is \(\frac12(2)(19)=19\,\text{J}\), which exceeds the initial \(1\,\text{J}\) by exactly the calculated \(18\,\text{J}\) of work.

Worked Example: Piecewise Force and Negative Work

A \(2\,\text{kg}\) particle starts from rest at \(x=0\,\text{m}\). Its net force is \(5\,\text{N}\) from \(x=0\) to \(x=2\,\text{m}\), then \(-3\,\text{N}\) from \(x=2\) to \(x=5\,\text{m}\). Find the total work and final speed.

On the first interval, the force and displacement point in the same direction:

$$ W_1=(5\,\text{N})(2\,\text{m})=10\,\text{J}. $$

On the second interval, the force points opposite the displacement:

$$ W_2=(-3\,\text{N})(5-2)\,\text{m}=-9\,\text{J}. $$

The total net work is \(W_{\text{net}}=W_1+W_2=10-9=1\,\text{J}\). Since the particle starts from rest, \(K_i=0\), so \(K_f=1\,\text{J}\). Then

$$ 1\,\text{J}=\frac12(2\,\text{kg})v_f^2, \qquad |v_f|=1\,\text{m/s}. $$

After the first interval, its kinetic energy is \(10\,\text{J}\), so it is still moving forward when it enters the second interval. The negative work removes \(9\,\text{J}\), leaving \(1\,\text{J}\); it does not stop before reaching \(x=5\,\text{m}\). Check: \(K_f-K_i=1-0=1\,\text{J}\), equal to the sum \(10+(-9)=1\,\text{J}\).

A Common Error: Using the Endpoint Force

A variable force cannot generally be treated as if its value at the endpoint acted over the entire displacement. For the force \(F(x)=(4\,\text{N/m})x\) from \(0\,\text{m}\) to \(3\,\text{m}\), one incorrect calculation is

$$ F(3)(3\,\text{m})=(12\,\text{N})(3\,\text{m})=36\,\text{J}. $$

This uses the final force throughout the interval, even though the force changes with position. The correct work is the integral \(W_{\text{net}}=\int_0^3(4\,\text{N/m})x\,dx=18\,\text{J}\). The endpoint-force calculation doubles the correct result in this case. For a constant force, multiplying force by displacement is valid; for a variable force, integrate unless a justified average force is being used.

Why the Work-Energy Theorem Is Useful

The theorem can answer questions about speed without first solving for the full motion as a function of time. If force is given as a function of position, an integral gives the work directly, and the change in kinetic energy follows. This is especially useful when a time-based description would be inconvenient.

The theorem also provides a sign check. Positive net work means kinetic energy increases; negative net work means it decreases; zero net work means initial and final kinetic energies are equal. Zero net work does not imply that no forces acted, only that their total work over the motion was zero.

Key takeaway: Add the signed work of all forces to obtain net work, then use \(W_{\text{net}}=K_f-K_i\). The result relates the initial and final speeds, but direction must be determined from the motion itself.

Check Your Understanding

Use the work-energy theorem and show the signed work in each calculation.

  1. A \(4\,\text{kg}\) object speeds up from \(2\,\text{m/s}\) to \(5\,\text{m/s}\). What is the net work done on it?
  2. A particle moves from \(x=1\,\text{m}\) to \(x=3\,\text{m}\) under \(F_{\text{net}}(x)=2x\,\text{N/m}\). Find the net work.
  3. Why does the derivation remain valid when the particle is momentarily at rest?
  4. If the net work is negative, what does that say about the initial and final kinetic energies?
  5. Why does the work-energy theorem alone not determine the sign of the final velocity?

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