Gravitational potential energy and kinetic energy

1. Overview

Energy is a scalar quantity that represents the capacity of a system to perform work. In the context of mechanics, the Principle of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one store to another. This topic focuses on two primary energy stores: Kinetic Energy ($E_K$), associated with the motion of a mass, and Gravitational Potential Energy ($E_P$), associated with the position of a mass within a uniform gravitational field. The fundamental link between these stores is Work Done ($W$), which is the mechanism by which energy is transferred. By applying Newton’s laws of motion and the definitions of work, we can derive the governing equations for these energy types, allowing for the analysis of complex physical systems where forces cause changes in speed and height.

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Key Definitions

To secure full marks in definition questions, the following precise terminology must be used:

• Gravitational Potential Energy ($E_P$): The energy stored in an object due to its position in a gravitational field. For AS Level, we specifically consider the change in this energy when an object is moved vertically in a uniform field.
• Kinetic Energy ($E_K$): The energy an object possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
• Work Done ($W$): The product of the force and the displacement moved in the direction of the force. It is measured in Joules (J), where $1 \, \text{J} = 1 \, \text{N m}$.
• Uniform Gravitational Field: A region where the gravitational field strength ($g$) is constant in both magnitude and direction at all points. This is a valid approximation for small changes in height near the surface of a planet.
• Acceleration of Free Fall ($g$): The acceleration of an object falling freely in a uniform gravitational field in the absence of air resistance. On Earth, this is taken as $9.81 \, \text{m s}^{-2}$.

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Content

3.1 Gravitational Potential Energy ($\Delta E_P$) in a Uniform Field

The formula $\Delta E_P = mg\Delta h$ describes the change in energy when an object of mass $m$ changes its vertical height by $\Delta h$ in a field of strength $g$.

Derivation of $\Delta E_P = mg\Delta h$ The syllabus requires you to derive this using the definition of work done ($W = Fs$).

1. Consider an object of mass $m$ being lifted vertically at a constant velocity.
2. To move the object at a constant velocity, the upward force $F$ applied must be exactly equal to the downward weight of the object to ensure there is no resultant force (Newton’s First Law). $$F = \text{Weight} = mg$$
3. The work done $W$ by the lifting force is defined as: $$W = F \times s$$ where $s$ is the displacement in the direction of the force.
4. In this scenario, the displacement $s$ is the change in vertical height, $\Delta h$.
5. Substituting the expression for force ($mg$) and displacement ($\Delta h$) into the work equation: $$W = (mg) \times \Delta h$$
6. Since the object is moving at a constant velocity, there is no change in kinetic energy. Therefore, all the work done by the external force is transferred into the gravitational potential energy store of the object. $$\Delta E_P = mg\Delta h$$

Important Constraints:

• This derivation assumes $g$ is constant. If the object moves significantly far from the Earth (e.g., into orbit), $g$ decreases, and this linear formula is no longer valid.
• $\Delta h$ must be the vertical displacement. Horizontal movement in a uniform gravitational field involves zero work done against gravity because the force (weight) is perpendicular to the displacement.

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3.2 Kinetic Energy ($E_K$)

Kinetic energy is the energy of a mass $m$ moving at velocity $v$. It is a scalar quantity, meaning it does not have a direction, and it is always positive (since $v$ is squared).

Derivation of $E_K = \frac{1}{2}mv^2$ The syllabus requires the derivation of this formula using the equations of motion (SUVAT).

1. Consider a constant resultant force $F$ acting on an object of mass $m$.
2. The object is initially at rest, so its initial velocity $u = 0$.
3. The force $F$ acts over a displacement $s$, accelerating the object to a final velocity $v$.
4. From the definition of work done: $$W = Fs$$
5. From Newton’s Second Law, the resultant force is the product of mass and acceleration: $$F = ma$$ Substituting this into the work equation gives: $$W = (ma)s = mas$$
6. We use the equation of motion that relates velocity, acceleration, and displacement: $$v^2 = u^2 + 2as$$
7. Since the object started from rest ($u = 0$): $$v^2 = 2as$$ Rearranging this to isolate the term $as$: $$as = \frac{v^2}{2}$$
8. Substitute this expression for $as$ back into the work equation ($W = m \times as$): $$W = m \left( \frac{v^2}{2} \right) = \frac{1}{2}mv^2$$
9. The work done on the object is equal to the kinetic energy $E_K$ it has gained. $$E_K = \frac{1}{2}mv^2$$

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3.3 The Principle of Conservation of Mechanical Energy

In a "closed" system where only conservative forces (like gravity) act and non-conservative forces (like air resistance or friction) are negligible, the total mechanical energy remains constant.

$$\text{Total Initial Energy} = \text{Total Final Energy}$$ $$(E_K + E_P)_{\text{initial}} = (E_K + E_P)_{\text{final}}$$

Case 1: An object falling from rest in a vacuum

• Loss in $E_P$ = Gain in $E_K$
• $mgh = \frac{1}{2}mv^2$
• $gh = \frac{1}{2}v^2 \implies v = \sqrt{2gh}$
• Note: The mass $m$ cancels out, proving that all objects fall at the same rate regardless of mass when air resistance is ignored.

Case 2: Systems with resistive forces If friction or air resistance is present, some energy is transferred to the surroundings as thermal energy (heat). $$\text{Loss in } E_P = \text{Gain in } E_K + \text{Work Done against friction}$$ $$mgh = \frac{1}{2}mv^2 + (f \times d)$$ where $f$ is the average resistive force and $d$ is the distance over which it acts.

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3.4 Worked Examples

Worked Example 1 — Motion on an Inclined Plane

A block of mass $450 \, \text{g}$ is pushed $3.50 \, \text{m}$ up a smooth slope inclined at $25.0^\circ$ to the horizontal. Calculate the change in the gravitational potential energy of the block.

Step 1: Convert units to SI. $m = 450 \, \text{g} = 0.450 \, \text{kg}$

Step 2: Determine the vertical height change ($\Delta h$). The displacement along the slope is $3.50 \, \text{m}$. Using trigonometry: $\Delta h = 3.50 \times \sin(25.0^\circ)$ $\Delta h = 1.479 \, \text{m}$

Step 3: Apply the GPE formula. $\Delta E_P = mg\Delta h$ $\Delta E_P = 0.450 \times 9.81 \times 1.479$ $\Delta E_P = 6.529... \, \text{J}$

Step 4: Round to appropriate significant figures (3 s.f.). Answer: $\Delta E_P = 6.53 \, \text{J}$

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Worked Example 2 — Energy Conversion with Resistive Forces

A child of mass $30.0 \, \text{kg}$ slides down a playground slide from a vertical height of $4.00 \, \text{m}$. The child reaches the bottom with a speed of $7.20 \, \text{m s}^{-1}$. Calculate the work done against friction during the descent.

Step 1: Calculate initial $E_P$ at the top. $E_P = mgh = 30.0 \times 9.81 \times 4.00$ $E_P = 1177.2 \, \text{J}$

Step 2: Calculate final $E_K$ at the bottom. $E_K = \frac{1}{2}mv^2 = 0.5 \times 30.0 \times (7.20)^2$ $E_K = 0.5 \times 30.0 \times 51.84$ $E_K = 777.6 \, \text{J}$

Step 3: Use the conservation of energy principle. Initial $E_P = \text{Final } E_K + \text{Work Done against friction } (W_f)$ $1177.2 = 777.6 + W_f$ $W_f = 1177.2 - 777.6$ $W_f = 399.6 \, \text{J}$

Step 4: Round to appropriate significant figures (3 s.f.). Answer: $W_f = 400 \, \text{J}$

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Key Equations

Quantity	Equation	Symbols	SI Units	Formula Sheet?
GPE Change	$\Delta E_P = mg\Delta h$	$m$: mass, $g$: $9.81 \, \text{m s}^{-2}$, $\Delta h$: vertical height	Joules (J)	Yes
Kinetic Energy	$E_K = \frac{1}{2}mv^2$	$m$: mass, $v$: speed/velocity	Joules (J)	Yes
Work Done	$W = Fs \cos \theta$	$F$: force, $s$: displacement, $\theta$: angle to force	Joules (J)	Yes
SUVAT	$v^2 = u^2 + 2as$	$u, v$: velocities, $a$: acceleration, $s$: displacement	$\text{m s}^{-1}, \text{m s}^{-2}, \text{m}$	Yes
Weight	$W = mg$	$m$: mass, $g$: $9.81 \, \text{m s}^{-2}$	Newtons (N)	No

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Common Mistakes to Avoid

• ❌ Wrong: Using $g = 9.8$ or $g = 10$.
• ✓ Right: Always use $g = 9.81 \, \text{m s}^{-2}$ as specified in the Cambridge Data Booklet.
• ❌ Wrong: Using the total distance traveled along a slope for $\Delta h$ in the $E_P$ formula.
• ✓ Right: Only the vertical component of displacement matters for gravitational potential energy. Use $s \sin \theta$ if necessary.
• ❌ Wrong: Forgetting to square the velocity in $\frac{1}{2}mv^2$ or forgetting the $\frac{1}{2}$ factor.
• ✓ Right: Write the formula out in full before substituting numbers to ensure no terms are missed.
• ❌ Wrong: Failing to convert mass from grams (g) to kilograms (kg).
• ✓ Right: Standard SI units are essential. $1 \, \text{kg} = 1000 \, \text{g}$.
• ❌ Wrong: Treating energy as a vector.
• ✓ Right: Energy is a scalar. If an object changes direction but maintains the same speed, its kinetic energy remains constant.

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Exam Tips

1. Derivation Precision: When asked to derive $\Delta E_P = mg\Delta h$, you must explicitly state that the object is moved at constant velocity (so $F = mg$) and that work done is $F \times s$. Skipping these logical steps will lose marks.
2. The $v^2$ Relationship: Be prepared for "ratio" questions. If the speed of a car triples, its kinetic energy increases by a factor of $3^2 = 9$.
3. Significant Figures: In the 9702 syllabus, your final answer should usually be given to the same number of significant figures as the data provided in the question (typically 2 or 3 s.f.). Never give an answer to 1 s.f.
4. Energy-Displacement Graphs:
   • A graph of $E_K$ against $v^2$ is a straight line through the origin with a gradient of $\frac{1}{2}m$.
   • A graph of $E_P$ against $h$ is a straight line through the origin with a gradient of $mg$.
5. Efficiency Links: While this topic focuses on $E_K$ and $E_P$, exams often link this to Power ($P = \frac{W}{t}$) or Efficiency. Always check if the question asks for the rate of energy transfer.
6. Standard Form: For very small energies (e.g., subatomic particles) or very large energies (e.g., vehicles), use standard form (e.g., $1.6 \times 10^{-19} \, \text{J}$) to avoid zero-counting errors.