Questions: Work-Energy Theorem: Rigorous Derivation and Applications

Question 1

A block slides down a frictionless ramp. You want to find its speed at the bottom. Which approach is most efficient, and why?

• A. Newton-Euler: sum all forces including the normal force, apply F=ma along the ramp direction
• B. Work-energy: only gravity does work (normal force is perpendicular to motion and does zero work), so W_gravity = ΔKE directly gives the final speed
• C. Both methods require exactly the same effort — work-energy only helps when friction is present
• D. Work-energy cannot be used here because the ramp is frictionless

Answer: (B)

The work-energy theorem's power is that forces perpendicular to motion do zero work and disappear from the calculation. On a frictionless ramp, the normal force is always perpendicular to velocity — it doesn't appear in the work-energy equation at all. Only gravity does work, so W_gravity = mgh = ΔKE = ½mv². Solving gives v = √(2gh) with no need to resolve forces along the ramp angle or compute the normal force. Newton-Euler works too, but requires more steps and careful geometry.

Question 2

A block slides across a rough horizontal surface and slows to a stop. You apply the work-energy theorem to find the stopping distance. Which forces must be included in the net work calculation?

• A. Only the normal force, since it balances gravity on a horizontal surface
• B. Only gravity, since friction is an internal force of the block-surface system
• C. Only kinetic friction, since it is the only force that does work (gravity and normal force are perpendicular to motion on a horizontal surface)
• D. All forces: gravity, normal force, and friction — the theorem requires net work from all forces

Answer: (C)

The net work theorem W_net = ΔKE includes all forces, but forces perpendicular to motion do zero work. On a horizontal surface: gravity acts downward (perpendicular to horizontal motion) — zero work. Normal force acts upward (perpendicular) — zero work. Only kinetic friction acts horizontally (opposing motion) — it does negative work W_friction = -f_k × d. Setting -f_k × d = 0 - ½mv₀² and solving gives the stopping distance. This is the correct and efficient application of the theorem.

Question 3

The work-energy theorem states that the net work done on a body equals its change in kinetic energy, and this result follows directly from integrating F = ma — no additional assumptions are required.

• T. True
• F. False

Answer: True

The derivation is direct: start with F = ma, dot both sides with velocity v to get F·v = ma·v = m(dv/dt)v = d(½mv²)/dt. Integrate over time: ∫F·v dt = Δ(½mv²), and since ∫F·v dt = ∫F·ds = W, the result is W_net = ΔKE. No assumptions about force type, path shape, or system properties were made — it is a pure mathematical consequence of Newton's second law. This is why the theorem is universally applicable to all force types.

Question 4

The work-energy theorem can seldom be directly applied to rigid bodies because internal forces between particles within the body complicate the calculation.

• T. True
• F. False

Answer: False

For a truly rigid body, internal forces come in Newton's third law pairs — equal and opposite — acting between particles that don't move relative to each other (by the rigid body assumption). These internal force pairs do zero net work, so they cancel out of the work-energy calculation. Only external forces remain. The theorem therefore applies cleanly to rigid bodies: W_net (external) = ΔKE. The complication only arises for deformable bodies or when internal energy changes (heat generation, plastic deformation) occur.

Question 5

Why is the work-energy approach often preferable to direct Newton-Euler force analysis, and in what situations does it provide the greatest advantage?

Model answer: Work-energy converts a vector problem into a scalar one: you only need to calculate the work done by each force (a dot product — one number), not resolve vector components at every point. Constraint forces perpendicular to motion (normal forces, tension in inextensible strings) automatically contribute zero work and drop out. The greatest advantage is when forces are position-dependent (springs: F = -kx), paths are curved, or constraint forces are unknown and irrelevant to the quantity you want.

The deeper reason is that work-energy is a scalar energy balance — kinetic energy is a scalar, work is a scalar integral. Finding the normal force in a roller-coaster problem requires careful geometry at every point; finding the speed at a specific height requires only the height difference and the conservation of energy. This is why energy methods are the first tool to reach for in dynamics problems involving speeds, heights, or springs, and why they become even more powerful in the Lagrangian formulation.