Edexcel A Level (IAL) Physics-1.18 - 1.19 Kinetic Energy & Gravitational Potential Energy- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -1.18 – 1.19 Kinetic Energy & Gravitational Potential Energy- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -1.18 – 1.19 Kinetic Energy & Gravitational Potential Energy- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 1.18 be able to use the equation $E_k = \frac{1}{2} mv^2$ for the kinetic energy of a body
- 1.19 be able to use the equation $\Delta E_{grav} = mg\Delta h$ for the difference in gravitational potential energy near the Earth’s surface
Kinetic Energy \( E_k = \tfrac{1}{2}mv^2 \)
Kinetic Energy is the energy a body has because of its motion.
Definition: The kinetic energy of a moving object depends on its mass and the square of its velocity.
\( E_k = \dfrac{1}{2}mv^2 \)
- \( E_k \) = kinetic energy (J)
- \( m \) = mass (kg)
- \( v \) = velocity (m s⁻¹)
Key Points:
- Kinetic energy is always positive (since \( v^2 \) is positive).
- If the velocity doubles, kinetic energy becomes four times greater.
- An object at rest has zero kinetic energy.
- Kinetic energy depends far more strongly on speed than on mass.
Understanding the Formula
- Derived from the work-energy theorem: the work done on an object equals its change in kinetic energy.
- Shows how force and displacement combine to produce motion.
- Useful in mechanics, collisions, braking problems, and energy transfer calculations.
Example (Easy)
Find the kinetic energy of a \( 3.0\, \mathrm{kg} \) object moving at \( 4\, \mathrm{m\,s^{-1}} \).
▶️ Answer / Explanation
\( E_k = \dfrac{1}{2}mv^2 = \dfrac{1}{2}(3.0)(4^2) = 1.5 \times 16 = 24\, \mathrm{J} \)
Example (Medium)
A \( 1200\, \mathrm{kg} \) car travels at \( 15\, \mathrm{m\,s^{-1}} \). Calculate its kinetic energy.
▶️ Answer / Explanation
\( E_k = \dfrac{1}{2}mv^2 = \dfrac{1}{2}(1200)(15^2) = 600 \times 225 = 135000\, \mathrm{J} \)
Example (Hard)
A ball of mass \( 0.20\, \mathrm{kg} \) increases its speed from \( 6\, \mathrm{m\,s^{-1}} \) to \( 12\, \mathrm{m\,s^{-1}} \). Calculate the change in kinetic energy.
▶️ Answer / Explanation
Initial KE:
\( E_{k,i} = \dfrac{1}{2}(0.20)(6^2) = 0.1 \times 36 = 3.6\, \mathrm{J} \)
Final KE:
\( E_{k,f} = \dfrac{1}{2}(0.20)(12^2) = 0.1 \times 144 = 14.4\, \mathrm{J} \)
Change in KE:
\( \Delta E_k = 14.4 – 3.6 = 10.8\, \mathrm{J} \)
Gravitational Potential Energy \( \Delta E_{\text{grav}} = mg\Delta h \)
Gravitational Potential Energy (GPE) is the energy an object has because of its position in a gravitational field.
Definition: The change in gravitational potential energy near the Earth’s surface is given by:
\( \Delta E_{\text{grav}} = mg\Delta h \)
- \( \Delta E_{\text{grav}} \) = change in gravitational potential energy (J)
- \( m \) = mass of object (kg)
- \( g \) = gravitational field strength (≈ \( 9.8\, \mathrm{N\,kg^{-1}} \))
- \( \Delta h \) = change in height (m)
Key Points:
- GPE increases if the object is raised (\( \Delta h > 0 \)).
- GPE decreases if the object is lowered (\( \Delta h < 0 \)).
- Near Earth’s surface, \( g \) is approximately constant.
- This formula assumes height is small compared to Earth’s radius.
- Change in GPE depends only on vertical movement, not path taken.
Understanding the Concept
- Raising an object requires work against gravity → energy stored as GPE.
- When an object falls, GPE converts to kinetic energy.
- In free fall (ignoring air resistance): \( \Delta E_{\text{grav}} = \Delta E_k \).
Example (Easy)
A \( 5.0\, \mathrm{kg} \) object is lifted by \( 2.0\, \mathrm{m} \). Calculate the increase in GPE.
▶️ Answer / Explanation
\( \Delta E_{\text{grav}} = mg\Delta h = 5.0 \times 9.8 \times 2.0 = 98\, \mathrm{J} \)
Example (Medium)
A \( 25\, \mathrm{kg} \) suitcase is lowered from a height of \( 1.5\, \mathrm{m} \) to the floor. Find the change in GPE.
▶️ Answer / Explanation
Height decreases → \( \Delta h = -1.5 \, \mathrm{m} \).
\( \Delta E_{\text{grav}} = mg\Delta h = 25 \times 9.8 \times (-1.5) = -367.5\, \mathrm{J} \)
Negative sign indicates energy lost.
Example (Hard)
A roller coaster car of mass \( 500\, \mathrm{kg} \) descends from a height of \( 32\, \mathrm{m} \) to \( 8\, \mathrm{m} \). Assuming no friction, calculate the change in GPE and the gain in KE.
▶️ Answer / Explanation
Change in height:
\( \Delta h = 8 – 32 = -24\, \mathrm{m} \)
Change in GPE:
\( \Delta E_{\text{grav}} = mg\Delta h = 500 \times 9.8 \times (-24) = -117600\, \mathrm{J} \)
The car loses \( 117600\, \mathrm{J} \) of GPE, which becomes KE.
Gain in KE:
\( \Delta E_k = +117600\, \mathrm{J} \) (energy conservation)