Work, Energy and Power · Lesson 4 of 16
Variable Forces and Work from Graphs
“Calculate signed areas and integrals, and derive the general work–energy theorem.”
• Build the work integral from small displacements. • Interpret signed force-position areas. • Handle variable resistance without assuming uniform acceleration.
Replace one large displacement by small pieces
A spring gets harder to stretch as extension increases. Using its final force for the entire extension overestimates the work. Divide the motion into many small displacements. On each piece the force is approximately constant, so the small work is force component multiplied by displacement. Adding the pieces gives a rectangular-area approximation. As the widths tend to zero, the sum becomes an integral.
For motion toward increasing x, areas above the x-axis contribute positive work and those below contribute negative work. Reversing limits reverses the integral for the same position-dependent force. For velocity-dependent resistance, the force itself changes when motion reverses; do not simply reuse a position-only graph. An average force over displacement can replace the integral, but it is not generally the average force over time.
A force-position graph is an energy map
Problem
A trunk is pulled 20 m. The pull is 100 N for the first 10 m, then decreases linearly to 50 N at 20 m. Friction is 50 N throughout. Find the work terms.
- 1.The first pull area is a rectangle: 100×10 = 1000 J.
- 2.The second is a trapezium: ½(100+50)×10 = 750 J. Thus Wpull = 1750 J.
- 3.Friction lies below the axis: Wfriction = −50×20 = −1000 J.
- 4.Net work is 750 J, so the trunk gains 750 J of kinetic energy. The pull at 20 m is 50 N, not zero.
Derive the theorem for a changing force
For a fixed mass in an inertial frame, differentiate K = ½mv² with respect to time. The chain rule gives dK/dt = mv dv/dt = Fnet v in one dimension. Since v dt = dx, this becomes dK = Fnet dx. Integrate between the endpoints to obtain the theorem without assuming constant acceleration. The vector derivation uses dK/dt = m v·a = Fnet·v.
Problem
A 1 kg block enters x = 0.10 m at 2 m/s. Until x = 2.01 m, resistance is Fx = −k/x with k = 0.5 J. Find its exit speed.
- 1.Initial kinetic energy is 2 J. The work is −k∫dx/x = −0.5 ln(2.01/0.10) J.
- 2.ln is the natural logarithm. The ratio of distances is dimensionless, as any logarithm’s argument must be.
- 3.Work ≈ −1.50036 J, so Kf ≈ 0.49964 J and vf = √(2Kf/m) ≈ 0.99964 m/s, approximately 1.00 m/s.
- 4.Check that K remains nonnegative before using the full interval. A formal negative K would mean the block stopped earlier and the assumed traversal was impossible.
Problem
A 0.5 kg particle has v = a x^(3/2), where a = 5 m^(−1/2) s⁻¹. Find net work from x = 0 to 2 m.
- 1.Use endpoint kinetic energies rather than first solving for force. At x = 0, v = 0. At x = 2 m, v² = a²x³ = 25×8 = 200 m²/s².
- 2.Net work = ½×0.5×200 = 50 J.
- 3.The endpoint relation answers the work question. An exact trajectory departing from x = 0 may require attention to initial conditions; no travel-time inference is needed here.
Check your understanding
Quiz
Area under an Fx–x graph gives
Explanation: Force multiplied by displacement has energy units.
If F rises linearly from 0 to 20 N over 4 m, its work is
Explanation: The triangular area is ½×4×20.
An integral predicts negative final K on a supposed route. This means
Explanation: The body stops before reaching that endpoint.
For ∫dx/x, the evaluated logarithm must involve
Explanation: Logarithms require dimensionless arguments.
Work–energy directly supplies
Explanation: It is a scalar equation and generally lacks timing and directional information.
Key Takeaways
• Variable-force work is a signed integral or area. • Derive ΔK = Wnet using the chain rule and Newton’s law. • Check reachability and distinguish distance averages from time averages.