CIE AS & A Level Physics 5.2 Gravitational potential energy and kinetic energy Study Notes- 2025-2027 Syllabus
CIE AS & A Level Physics 5.2 Gravitational potential energy and kinetic energy Study Notes – New Syllabus
CIE AS & A Level Physics 5.2 Gravitational potential energy and kinetic energy Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics Study Notes syllabus with Candidates should be able to:
- derive, using W = Fs, the formula ΔEP = mgΔh for gravitational potential energy changes in a uniform gravitational field
- recall and use the formula ΔEP = mgΔh for gravitational potential energy changes in a uniform gravitational field
- derive, using the equations of motion, the formula for kinetic energy \(E_k=\frac{1}{2}mv^2\)
- recall and use \(E_k=\frac{1}{2}mv^2\)
Derivation of the Formula ( \mathrm{\Delta E_P = mg\Delta h} ) Using ( \mathrm{W = Fs} )
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field.
Derivation:
Consider an object of mass \( \mathrm{m} \) being raised vertically through a height \( \mathrm{\Delta h} \) in a uniform gravitational field of strength \( \mathrm{g.} \)
The weight of the object is:
\( \mathrm{W = mg} \)
Work is done against the force of gravity to raise the object through a vertical displacement \( \mathrm{s = \Delta h.} \)
From the definition of work done:
\( \mathrm{Work\ done = Force \times Distance\ moved\ in\ direction\ of\ force} \)
\( \mathrm{W = F s} \)
Since the force applied must balance the weight to lift the object at constant speed:
\( \mathrm{F = mg} \)
Therefore:
\( \mathrm{W = mg\Delta h} \)
This work done is stored as an increase in gravitational potential energy:
\( \mathrm{\Delta E_P = mg\Delta h} \)
Hence,
\( \mathrm{\Delta E_P = mg\Delta h} \)
- \( \mathrm{\Delta E_P} \): change in gravitational potential energy (J)
- \( \mathrm{m} \): mass of the object (kg)
- \( \mathrm{g} \): gravitational field strength (m·s⁻²)
- \( \mathrm{\Delta h} \): change in height (m)
Physical Meaning:
- Raising an object increases its potential energy — energy is stored due to work done against gravity.
- Lowering an object decreases its potential energy — energy is released as the object does work on its surroundings.
- In a uniform field (e.g. near Earth’s surface), \( \mathrm{g} \) is constant, so the change in GPE is directly proportional to the height change.
Example
Explain why the gain in gravitational potential energy when lifting a mass vertically at constant velocity equals the work done by the lifting force.
▶️ Answer / Explanation
At constant velocity, the lifting force equals the weight \( \mathrm{(F = mg).} \) The work done by the force in moving the object through height \( \mathrm{\Delta h} \) is \( \mathrm{W = F\Delta h = mg\Delta h.} \) This energy is stored as an increase in gravitational potential energy, so \( \mathrm{\Delta E_P = mg\Delta h.} \)
Use of the Formula ( \mathrm{\Delta E_P = mg\Delta h} )
Applications:
- Calculating the energy required to lift objects to a certain height.
- Determining potential energy changes in pendulums, projectiles, or falling bodies.
- Analyzing energy conservation in mechanical systems.
Sign Convention:
- If the object is raised, \( \mathrm{\Delta h > 0} \) → \( \mathrm{\Delta E_P > 0} \) (gain in energy).
- If the object is lowered, \( \mathrm{\Delta h < 0} \) → \( \mathrm{\Delta E_P < 0} \) (loss in energy).
Gravitational Potential Energy
| Quantity | Expression | SI Unit | Description |
|---|---|---|---|
| Work Done | \( \mathrm{W = Fs} \) | J (joule) | Energy transferred when a force moves through a distance |
| Gravitational Potential Energy Change | \( \mathrm{\Delta E_P = mg\Delta h} \) | J (joule) | Energy gained or lost due to a change in height in a uniform field |
Key Takeaways:
- Work done against gravity equals the gain in gravitational potential energy.
- In a uniform field, \( \mathrm{g} \) is constant, so GPE changes linearly with height.
- \( \mathrm{\Delta E_P = mg\Delta h} \) is valid near Earth’s surface or any region where \( \mathrm{g} \) is approximately constant.
Example
A 5.0 kg mass is lifted vertically through a height of \( \mathrm{2.5\,m.} \) Calculate the gain in gravitational potential energy. Take \( \mathrm{g = 9.8\,m·s^{-2}.} \)
▶️ Answer / Explanation
Using \( \mathrm{\Delta E_P = mg\Delta h} \):
\( \mathrm{\Delta E_P = (5.0)(9.8)(2.5) = 122.5\,J.} \)
The mass gains \( \mathrm{1.23\times10^2\,J} \) of gravitational potential energy.
Example
A 2.0 kg rock falls from a height of \( \mathrm{8.0\,m.} \) Calculate the change in gravitational potential energy as it falls. Take \( \mathrm{g = 9.8\,m·s^{-2}.} \)
▶️ Answer / Explanation
Using \( \mathrm{\Delta E_P = mg\Delta h} \):
\( \mathrm{\Delta E_P = (2.0)(9.8)(-8.0) = -157\,J.} \)
The negative sign indicates a loss of potential energy as the rock falls — this energy is converted into kinetic energy (ignoring air resistance).
Derivation of the Kinetic Energy Formula
To derive the expression \( \mathrm{E_k = \tfrac{1}{2}mv^2} \) using the equations of motion and the work–energy principle.
Work–Energy Principle:
The work done on an object by a resultant force equals the change in its kinetic energy.
\( \mathrm{W = \Delta E_k} \)
Derivation Steps:
- Consider an object of mass \( \mathrm{m} \) moving in a straight line under a constant force \( \mathrm{F.} \)
- From Newton’s Second Law, \( \mathrm{F = ma.} \)
- If the object moves a distance \( \mathrm{s} \) while its velocity changes from \( \mathrm{u} \) to \( \mathrm{v,} \) the work done by the force is:
\( \mathrm{W = F s.} \)
- Using \( \mathrm{F = ma} \), we get \( \mathrm{W = ma s.} \)
- From the equation of motion (for constant acceleration):
\( \mathrm{v^2 = u^2 + 2as.} \)
Rearrange for \( \mathrm{as} \):
\( \mathrm{as = \tfrac{1}{2}(v^2 – u^2).} \)
Substitute this into \( \mathrm{W = ma s:} \)
\( \mathrm{W = m \times \tfrac{1}{2}(v^2 – u^2)} \)
\( \mathrm{W = \tfrac{1}{2}m(v^2 – u^2)} \)
The work done increases the object’s kinetic energy, so:
\( \mathrm{E_k = \tfrac{1}{2}mv^2.} \)
Hence,
\( \mathrm{E_k = \tfrac{1}{2}mv^2} \)
- \( \mathrm{E_k} \): kinetic energy (J)
- \( \mathrm{m} \): mass of the object (kg)
- \( \mathrm{v} \): speed of the object (m·s⁻¹)
Physical Interpretation:
- Kinetic energy is the energy an object possesses due to its motion.
- It depends on both mass and the square of velocity — doubling the speed quadruples the kinetic energy.
- Work done by a resultant force increases an object’s kinetic energy by the same amount.
Example
A car of mass \( \mathrm{1000\,kg} \) accelerates uniformly from rest under a constant force of \( \mathrm{2000\,N} \) through a distance of \( \mathrm{50\,m.} \) Find its final speed using the work–energy principle.
▶️ Answer / Explanation
Work done on the car = change in kinetic energy:
\( \mathrm{W = F s = 2000 \times 50 = 1.0\times10^5\,J.} \)
Since \( \mathrm{W = \tfrac{1}{2}mv^2,} \)
\( \mathrm{1.0\times10^5 = \tfrac{1}{2}(1000)v^2} \)
\( \mathrm{v^2 = 200 \Rightarrow v = 14.1\,m/s.} \)
The car’s final speed is \( \mathrm{14.1\,m/s.} \)
Recall and Use of the Kinetic Energy Formula
Formula:
\( \mathrm{E_k = \tfrac{1}{2}mv^2} \)
Meaning:
Kinetic energy is the work required to accelerate a stationary object of mass \( \mathrm{m} \) to a velocity \( \mathrm{v.} \)
Key Observations:
- \( \mathrm{E_k} \propto v^2 \): doubling the speed increases kinetic energy fourfold.
- \( \mathrm{E_k} \propto m \): for the same speed, a heavier object has more kinetic energy.
- Measured in joules (J), where \( \mathrm{1\,J = 1\,kg·m^2·s^{-2}.} \)
Example
A ball of mass \( \mathrm{0.5\,kg} \) moves with a velocity of \( \mathrm{6.0\,m/s.} \) Calculate its kinetic energy. If the ball’s speed doubles, what happens to its kinetic energy?
▶️ Answer / Explanation
Using \( \mathrm{E_k = \tfrac{1}{2}mv^2} \):
\( \mathrm{E_k = \tfrac{1}{2}(0.5)(6.0)^2 = 0.25 \times 36 = 9.0\,J.} \)
If the speed doubles (\( \mathrm{v = 12.0\,m/s} \)),
\( \mathrm{E_k’ = \tfrac{1}{2}(0.5)(12.0)^2 = 0.25 \times 144 = 36\,J.} \)
Thus, when the speed doubles, kinetic energy becomes four times greater.