Work Energy and Power in Depth

Work, Energy, and Power in Depth

Further Mechanics extends energy methods beyond the basics, covering variable forces, elastic potential energy, and the work-energy theorem in more challenging contexts.

Work Done by a Constant Force

W = F · d = Fd cos α

where α is the angle between the force and the direction of motion.

Scenario	Work done
Force along direction of motion (α = 0)	W = Fd (positive)
Force opposite to motion (α = π)	W = −Fd (negative)
Force perpendicular to motion (α = π/2)	W = 0

Work Done by a Variable Force

If the force F varies with position x:

W = ∫_a^b F(x) dx

This is the area under the force-displacement graph.

Example: A force F = 3x² + 2 N acts on a particle moving from x = 1 to x = 4.

W = ∫₁^4 (3x² + 2) dx = [x³ + 2x]₁^4 = (64 + 8) − (1 + 2) = 69 J

Hooke's Law and Elastic Potential Energy

A spring or elastic string with natural length l and modulus of elasticity λ, extended by x:

Tension: T = λx/l (Hooke's law)

The elastic potential energy (EPE) stored:

EPE = λx²/(2l)

Derivation: EPE = ∫₀^x (λt/l) dt = λx²/(2l)

Important: x is the extension (or compression) beyond the natural length, not the total length.

For a string (can only be stretched): x ≥ 0.

For a spring (can be stretched or compressed): x can be positive or negative, but EPE is always positive.

The Work-Energy Theorem

The total work done on a particle equals its change in kinetic energy:

W_total = ΔKE = ½mv² − ½mu²

When multiple forces act, you can either:

1. Find the resultant force and integrate, or

2. Sum the work done by each force separately

Conservation of Energy (Extended)

When only conservative forces act (gravity, elastic forces):

KE + GPE + EPE = constant

½mv² + mgh + λx²/(2l) = constant

Example: A ball of mass 0.5 kg is attached to an elastic string (natural length 2 m, λ = 40 N) fixed to a ceiling. Released from the ceiling. Find the maximum extension.

At release (point A): KE = 0, GPE = 0 (take as datum), EPE = 0.

At maximum extension (point B, distance d below ceiling, extension x = d − 2):

• KE = 0 (momentarily at rest)
• GPE = −0.5 × 9.8 × d = −4.9d
• EPE = 40(d − 2)²/(2 × 2) = 10(d − 2)² (only if d > 2)

Conservation: 0 = −4.9d + 10(d − 2)²

10d² − 40d + 40 − 4.9d = 0 → 10d² − 44.9d + 40 = 0

Solve: d = (44.9 ± √(44.9² − 1600))/20 = (44.9 ± √(2016.01 − 1600))/20 = (44.9 ± √416.01)/20

d = (44.9 ± 20.40)/20 → d = 3.265 m or d = 1.225 m

Since d > 2 (string must be extended): d ≈ 3.27 m, extension ≈ 1.27 m

Power

Power = rate of doing work:

P = dW/dt = F · v = Fv cos α

For a vehicle with driving force F, resistance R, at velocity v:

• Acceleration phase: F − R = ma, so P = Fv
• Maximum velocity: F = R (no acceleration), so P = Rv_max

Example: A car of mass 1200 kg has engine power 36 kW. Resistance = 800 N. Find maximum speed and acceleration at 20 m/s.

Maximum speed: P = Rv_max → 36000 = 800 × v_max → v_max = 45 m/s

At 20 m/s: F = P/v = 36000/20 = 1800 N

ma = F − R = 1800 − 800 = 1000 N → a = 1000/1200 = 5/6 m/s²

Work-Energy with Friction

Friction is a non-conservative force — it dissipates energy as heat:

KE₁ + GPE₁ + EPE₁ = KE₂ + GPE₂ + EPE₂ + W_friction

where W_friction = μRd (friction force × distance).

Variable Resistance

If resistance depends on velocity (e.g., air resistance R = kv²):

Using Newton's second law: F − kv² = m(dv/dt)

This gives a separable ODE. At terminal velocity, acceleration = 0: F = kv_terminal²

Energy in Connected Particle Systems

For pulleys, inclined planes, and tow-bar problems, apply conservation of energy to the whole system:

Total work by external forces = Change in total KE + Change in total PE + Energy lost to friction

This avoids having to find tension forces explicitly.

Exam Tips

• Choose your datum for GPE wisely — usually the lowest point the object reaches
• EPE = λx²/(2l) uses extension x, not the total length — this is the most common error
• For elastic strings, the string goes slack when the extension is zero — the formula doesn't apply for compression
• Power = Fv is an instantaneous relationship — F and v must be at the same instant
• In energy equations, list every form of energy at each position — missing one term is the typical exam slip
• When a variable force depends on position, integrate for work; when it depends on velocity, you may need F = ma = mv(dv/dx)