Gravitational Potential Energy (GPE) & Kinetic Energy (KE)

1. The Work‑Energy Theorem – foundation for all energy concepts

• Work done by a constant force F acting through a displacement s in the direction of the force

  \$W = F\,s\$

• If the force is the weight of a body (F = mg) and the displacement is purely vertical (s = \Delta h)

  \$W_{\text{gravity}} = mg\,\Delta h\$

• Work‑energy theorem (Cambridge syllabus 5.1)

  \$W{\text{net}} = \Delta EK = \tfrac12 m vf^{2}-\tfrac12 m vi^{2}\$

• When gravity is the only force doing work, the work can be expressed as the negative change in a new form of energy – gravitational potential energy (GPE)

  \$\Delta EP \equiv -W{\text{gravity}} = -\,mg\,\Delta h\$

• Re‑arranging gives the textbook formula used throughout the syllabus (5.2)

  \$\boxed{\Delta E_P = mg\,\Delta h}\$

  (positive Δh → GPE increases; negative Δh → GPE decreases).

2. Gravitational Potential Energy

2.1 Definition and reference level

GPE is the energy a body possesses because of its position in a uniform gravitational field. It is defined relative to an arbitrarily chosen reference level where the potential energy is set to zero.

• Typical choice: ground level, h = 0 → E_P = 0.
• Only differences in GPE (ΔE_P) have physical meaning – the absolute value depends on the chosen reference.

2.2 Symbol, units and sign convention

2.3 Derivation of the formula

Starting from the work done by gravity (a constant, conservative force):

\[

W{\text{gravity}} = \int{hi}^{hf} (-mg)\,dh = -mg\,(hf-hi) = -mg\,\Delta h

\]

Because the work done by a conservative force equals the negative change in its associated potential energy,

\[

\Delta EP = -W{\text{gravity}} = mg\,\Delta h

\]

2.4 Power and efficiency (syllabus 5.4)

• Power is the rate at which work is done or energy is transferred:

  \$P = \frac{W}{t} = \frac{\Delta E}{\Delta t}\$

• When a person lifts a load at constant speed, the required power is

  \$P = \frac{mg\Delta h}{t}\$

• Efficiency compares useful mechanical energy output with the total energy supplied:

  \$\eta = \frac{\text{useful work output}}{\text{total work input}}\times100\%\$

3. Kinetic Energy – derivation from the work‑energy theorem (syllabus 5.3)

• For a particle of mass m accelerating from rest under a constant net force F, the work done is

  \$W = F\,s = m a s\$

• Using the kinematic relation \(v^{2}=2as\) (valid for constant acceleration), substitute \(as = \tfrac12 v^{2}\) to obtain

  \$W = \tfrac12 m v^{2}\$

• Since the work‑energy theorem tells us that this work equals the change in kinetic energy, the kinetic energy of a body moving with speed v is

  \[

  \boxed{E_K = \tfrac12 m v^{2}}

  \]

4. Mechanical Energy Conservation (syllabus 5.1)

When only conservative forces (gravity, ideal springs) do work, the total mechanical energy remains constant:

\[

E{\text{mech}} = EP + E_K = \text{constant}

\]

\[

E{P,i}+E{K,i}=E{P,f}+E{K,f}

\]

4.1 When the statement is valid

• No non‑conservative forces (friction, air resistance, tension in a non‑ideal rope) act.
• All external work can be represented by a conservative potential.

4.2 When mechanical energy is not conserved

• Presence of friction, drag, or any dissipative force.
• Energy is transferred to heat, sound, internal energy, etc.
• In such cases use the full work‑energy theorem, including non‑conservative work:

  \[

  \Delta EK = W{\text{gravity}} + W_{\text{nc}}

  \]

  where \(W_{\text{nc}}\) is the net work of non‑conservative forces.

5. Extending the Energy Theme to Other Syllabus Areas

5.1 Elastic (spring) potential energy – syllabus 6

For an ideal spring obeying Hooke’s law (\(F = -kx\)) the stored energy is

\[

E_{\text{spring}} = \tfrac12 k x^{2}

\]

This follows the same energy‑conservation logic used for GPE and KE and will be needed in later questions (e.g., vertical spring‑mass systems).

5.2 Energy in waves – syllabus 7

Mechanical waves transport energy. The average intensity (energy transferred per unit area per unit time) is proportional to the square of the amplitude:

\[

I \propto A^{2}

\]

While not directly part of the GPE/KE topic, recognising that “energy can be carried by a wave” helps students compare mechanical‑energy ideas with wave‑energy concepts later in the course.

5.3 Electrical energy – syllabus 9

Energy transferred by electric charges is

\[

W = QV \quad\text{or}\quad E = \tfrac12 C V^{2}

\]

Both expressions are analogous to the mechanical work‑energy relation and illustrate that the same conservation principles apply across domains.

5.4 Nuclear energy – syllabus 11

Mass–energy equivalence (Einstein’s \(E = mc^{2}\)) introduces a different form of stored energy – nuclear binding energy. It is useful to remind students that gravitational potential energy is a mechanical form, distinct from the nuclear form that appears later.

6. Systematic Procedure for Solving Exam Questions

1. Read the question carefully. Identify:

   • Known quantities (mass, height, speed, angle, time, power, etc.).
   • What is required (ΔE_P, final speed, maximum height, power, efficiency, etc.).
   • Any non‑conservative forces mentioned.

2. Choose a convenient reference level for GPE. Usually the lowest point in the problem is taken as \(h=0\) (so \(E_P=0\) there).
3. Write down the relevant equations.

   • ΔE_P = mg Δh
   • E_K = ½ mv²
   • E_{\text{spring}} = ½ kx² (if a spring is involved)
   • Mechanical‑energy conservation (if only conservative forces act)
   • Power: \(P = \frac{\Delta E}{\Delta t}\)
   • Efficiency: \(\eta = \frac{\text{useful work}}{\text{total work}}\times100\%\)

4. Apply the sign convention for Δh.

   \(\Delta h = h{\text{final}}-h{\text{initial}}\).

   Positive → upward displacement; negative → downward.

5. Substitute numbers, keep units consistent. Use \(g = 9.81\ \text{m s}^{-2}\) unless the question specifies otherwise (sometimes 10 m s⁻² is allowed for quick estimates).
6. Solve algebraically before inserting numbers. This reduces rounding errors.
7. Check the answer.

   • Correct units (J for energy, W for power, m s⁻¹ for speed).
   • Reasonable magnitude (e.g., a 2 kg object dropped 5 m should have a speed ≈10 m s⁻¹).
   • Sign of the result matches the physical situation.
   • If friction or other non‑conservative forces are present, verify that their work has been accounted for.

7. Worked Examples

Example 1 – Simple vertical drop (mechanical‑energy conservation)

Problem: A 2.0 kg ball is released from rest at a height of 5.0 m. Find its speed just before it reaches the ground (ignore air resistance).

1. Reference level: ground (\(h=0\)).
2. Initial GPE: \(E_{P,i}=mg h = (2.0)(9.81)(5.0)=98.1\ \text{J}\).
3. Initial KE: \(E_{K,i}=0\).
4. At the ground: \(E_{P,f}=0\); let final speed be \(v\).
5. Conservation of mechanical energy:

   \(E{P,i}+E{K,i}=E{P,f}+E{K,f}\;\Rightarrow\;98.1 = \tfrac12 (2.0)v^{2}\).

6. Solve: \(v = \sqrt{\frac{2\times98.1}{2.0}} \approx 9.9\ \text{m s}^{-1}\).

Example 2 – Lifting a crate (use of ΔE_P and power)

Problem: A 5.0 kg crate is lifted vertically upward by 3.2 m at constant speed in 4.0 s. Calculate (a) the increase in GPE and (b) the average power supplied.

1. Δh = +3.2 m → \(\Delta E_P = mg\Delta h = (5.0)(9.81)(3.2)=156.96\ \text{J}\).
2. Because the speed is constant, net work on the crate is zero; the work done by the person equals the increase in GPE.
3. Average power: \(P = \frac{W}{t} = \frac{156.96\ \text{J}}{4.0\ \text{s}} = 39.2\ \text{W}\).

Example 3 – Object on a frictionless incline

Problem: A 1.2 kg block slides down a smooth incline that is 4.0 m high. Find its speed at the bottom.

1. Reference level: bottom (\(h=0\)).
2. Initial GPE: \(E_{P,i}=mg h = (1.2)(9.81)(4.0)=47.1\ \text{J}\).
3. Initial KE: 0.
4. At the bottom: \(E_{P,f}=0\); let final speed be \(v\).
5. Energy conservation (no friction): \(mg h = \tfrac12 m v^{2}\) → \(v = \sqrt{2gh}= \sqrt{2(9.81)(4.0)}\approx 8.86\ \text{m s}^{-1}\).

Example 4 – Projectile reaching its maximum height

Problem: A 0.50 kg stone is thrown vertically upward with an initial speed of 12 m s⁻¹. Determine the maximum height above the launch point.

1. At the highest point \(E_K = 0\).
2. Initial KE: \(\tfrac12 m v^{2}= \tfrac12 (0.50)(12^{2}) = 36\ \text{J}\).
3. All this energy becomes GPE: \(mg\Delta h = 36\ \text{J}\).
4. \(\Delta h = \frac{36}{(0.50)(9.81)} \approx 7.34\ \text{m}\).

Example 5 – Spring‑mass system with gravity (syllabus 6)

Problem: A 0.80 kg mass is attached to a vertical spring (force constant \(k = 250\ \text{N m}^{-1}\)). The spring is initially uncompressed and the mass is released from rest. Find the maximum compression of the spring.

1. At the lowest point the speed is zero, so total mechanical energy is purely elastic + gravitational:

   \(\tfrac12 k x^{2} = mgx\) where \(x\) is the compression (downward positive).

2. Rearrange: \(\tfrac12 k x^{2} - mgx = 0 \Rightarrow x\bigl(\tfrac12 k x - mg\bigr)=0\).
3. Discard the trivial solution \(x=0\); solve \(\tfrac12 k x = mg\):

   \(x = \frac{2mg}{k}= \frac{2(0.80)(9.81)}{250}\approx 0.063\ \text{m}=6.3\ \text{cm}\).

Example 6 – Power required to raise a load (syllabus 5.4)

Problem: A 150 kg crate is lifted 10 m in 5 s at constant speed. Determine the average power output of the lifting mechanism.

1. ΔE_P = mgΔh = (150)(9.81)(10) = 14 715 J.
2. Power: \(P = \frac{ΔE}{Δt} = \frac{14 715\ \text{J}}{5\ \text{s}} = 2 943\ \text{W}\) (≈ 2.94 kW).
3. If the mechanism is 80 % efficient, the input power required is \(\frac{P}{0.80} \approx 3.68\ \text{kW}\).

8. Common Mistakes & How to Avoid Them

• Sign of Δh: Remember \(\Delta h = h{\text{final}}-h{\text{initial}}\). Upward motion → positive Δh; downward → negative.
• Confusing work and energy: Work done by gravity is \(-mg\Delta h\); the change in GPE is the opposite sign.
• Using the wrong value of g: Follow the exam instruction (9.81 m s⁻², 9.80 m s⁻², or the rounded 10 m s⁻²).
• Omitting the mass factor: Both ΔEP and EK contain the mass; dropping it leads to a factor‑of‑m error.
• Assuming energy conservation when friction is present: Identify any non‑conservative forces first; include their work explicitly.
• Unit errors: Energy → joules (kg·m²·s⁻²); power → watts (J s⁻¹); speed → m s⁻¹; height → metres.
• Forgetting efficiency: When a machine is involved, state whether the answer required is the useful output or the total input.

9. Summary Checklist

1. Read the question → list knowns and unknowns.
2. Choose a reference level for GPE (usually the lowest point).
3. Write down the relevant equations:

   • ΔE_P = mg Δh
   • E_K = ½ mv²
   • E_{\text{spring}} = ½ kx² (if applicable)
   • Mechanical‑energy conservation (if only conservative forces act)
   • Power: \(P = ΔE/Δt\)
   • Efficiency: \(\eta = \frac{\text{useful work}}{\text{total work}}\times100\%\)

4. Apply the correct sign to Δh.
5. Substitute numbers, keep units consistent.
6. Solve algebraically before inserting numbers to minimise rounding.
7. Check:

   • Units (J for energy, W for power, m s⁻¹ for speed).
   • Reasonableness of magnitude.
   • That any non‑conservative work has been accounted for.
   • Efficiency and power values where required.

10. Practice Questions (with marks)

1. (2 marks) A 5.0 kg crate is lifted vertically by 3.2 m. Calculate the increase in its gravitational potential energy. (Use \(g = 9.81\ \text{m s}^{-2}\).)
2. (3 marks) A 150 kg crate is raised 10 m in 5 s at constant speed. Find the average power output of the lifting mechanism. Assume 100 % efficiency.
3. (4 marks) A 0.80 kg mass is attached to a vertical spring with \(k = 250\ \text{N m}^{-1}\). The spring is initially uncompressed and the mass is released from rest. Determine the maximum compression of the spring.
4. (5 marks) A 2.0 kg ball is thrown upward from ground level with an initial speed of 15 m s⁻¹. Ignoring air resistance:

   • Find the maximum height reached.
   • Find the speed when the ball has descended back to a height of 5 m above the ground.

5. (6 marks) A smooth inclined plane rises 3.0 m vertically and is 5.0 m long. A 1.5 kg block starts from rest at the top and slides down the plane. The coefficient of kinetic friction is 0.10.

   • Calculate the speed of the block at the bottom.
   • Determine the total work done by friction.
   • State the mechanical energy of the block at the bottom.

11. Quick Reference Table