Work, Energy, Power and Conservation Laws

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your NECO exam.

Work ($W$): Energy transferred by a force. $W = F \cdot d \cdot \cos\theta$, where $\theta$ is the angle between the force and displacement. Unit: Joules (J).

• If force and displacement are in the same direction ($\theta = 0°$): $W = Fd$
• If perpendicular ($\theta = 90°$): $W = 0$ (e.g., centripetal force does no work)

Kinetic Energy ($KE$): $KE = \frac{1}{2}mv^2$

Potential Energy ($PE$): $PE = mgh$ (gravitational, near Earth’s surface)

Work–Energy Theorem: $W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Power ($P$): Rate of doing work. $P = \frac{W}{t} = Fv$. Unit: Watts (W).

Conservation of Mechanical Energy: In the absence of non-conservative forces (friction, air resistance): $$\frac{1}{2}mv^2 + mgh = \text{constant}$$

⚡ NECO Tip: When an object slides down an inclined plane, the work done by gravity $= mg \sin\theta \times d$ along the plane. Friction does negative work $= \mu R d = \mu mg \cos\theta \cdot d$. The net work = change in kinetic energy.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for NECO Physics students with a few days to months.

Gravitational Potential Energy (General): $PE = \dfrac{GMm}{r}$ (at distance $r$ from Earth’s centre, where $M$ is Earth’s mass)

Elastic Potential Energy: $PE_{\text{spring}} = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is extension.

Spring Force (Hooke’s Law): $F = kx$ (for small extensions)

Efficiency: $$\eta = \frac{\text{useful output energy}}{\text{input energy}} \times 100% = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100%$$

Conservation of Energy with Friction:

When friction acts: $W_{\text{by friction}} = \Delta KE + \Delta PE = -\mu R \cdot d$ The energy lost to friction appears as heat.

Example: A 2 kg block slides down a rough inclined plane (angle $30°$, coefficient of friction $\mu = 0.2$) over a distance of $5$ m. Find its speed at the bottom.

• Work by gravity: $mg\sin 30° \times 5 = 2(10)(0.5)(5) = 50$ J
• Work by friction: $-\mu mg\cos 30° \times 5 = -0.2(2)(10)(0.866)(5) = -17.32$ J
• Net work = $50 - 17.32 = 32.68$ J = $\frac{1}{2}(2)v^2 \Rightarrow v = \sqrt{32.68} = 5.72$ m/s

Power and Velocity: $P = Fv$ is useful when force and velocity are constant. For variable situations: $P = \dfrac{dW}{dt}$.

Kinetic Energy and Momentum: $KE = \dfrac{p^2}{2m}$ and $p = mv$, so $KE = \frac{1}{2}mv^2 = \frac{p^2}{2m}$.

⚡ NECO Common Mistakes:

• Forgetting that work done by friction is always negative
• Mixing up $mgh$ (gravitational PE near Earth’s surface) with $G Mm/r$ (general gravitational PE)
• In power questions, using average force × average velocity when the question specifies instantaneous or constant values
• Confusion between energy (scalar, J) and power (rate, W)

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for NECO and JAMB Physics preparation.

Derivation of Kinetic Energy from Newton’s Second Law:

Starting from $F = ma$ and $v^2 = u^2 + 2as$: $$W = \int \mathbf{F} \cdot d\mathbf{r} = \int m\frac{dv}{dt}, dr = \int mv, dv = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$ This confirms the work–energy theorem.

Derivation of Gravitational Potential Energy:

Work done against gravity to lift mass $m$ through height $h$: $$W = \int_0^h mg, dy = mgh$$

Conservative vs Non-Conservative Forces:

• Conservative: Work done is path-independent. Gravity, elastic force. Total mechanical energy is conserved.
• Non-conservative: Work done depends on path. Friction, air resistance. Mechanical energy is not conserved.

The Work–Energy Principle for Variable Force: $$W = \int_{r_1}^{r_2} F, dr = \Delta KE$$

Potential Energy from a Force: $$U(x) = -\int \mathbf{F} \cdot d\mathbf{r} + \text{constant}$$

Collisions and Conservation Laws:

• Elastic collision: Both momentum AND kinetic energy are conserved. $v_1’ = \frac{(m_1 - m_2)v_1 + 2m_2v_2}{m_1+m_2}$
• Inelastic collision: Only momentum conserved. Kinetic energy is lost.
• Perfectly inelastic collision: Objects stick together. Maximum KE is lost.

Example (elastic collision): A 1 kg ball moving at 5 m/s collides elastically with a stationary 2 kg ball. Find their speeds after collision.

• $v_1’ = \frac{(1-2)(5) + 2(2)(0)}{3} = \frac{-5}{3} = -1.67$ m/s (reverses direction)
• $v_2’ = \frac{2(1)(5) + (2-2)(0)}{3} = \frac{10}{3} = 3.33$ m/s

Power in Fluids: For fluid flow: $P = \rho g Q h$, where $\rho$ = density, $Q$ = discharge rate (m³/s), $h$ = height.

NECO/JAMB Question Patterns:

• NECO often asks: find speed at bottom of incline using energy conservation; calculate power given force and velocity; distinguish between elastic and inelastic collisions
• Watch for questions involving roller coasters, pendulums, or projectile motion combined with energy

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