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In this lesson you will learn how to calculate the kinetic energy of moving objects and the gravitational potential energy of objects at height. These two energy stores are among the most commonly examined calculations in the AQA GCSE Physics specification (Section 4.1).
Kinetic energy is the energy stored in the movement of an object. Any object that is moving has kinetic energy. The faster it moves and the more massive it is, the more kinetic energy it has.
The kinetic energy of a moving object can be calculated using:
E_k = 0.5 x m x v^2
Where:
| Quantity | Symbol | Unit |
|---|---|---|
| Kinetic energy | E_k | joules (J) |
| Mass | m | kilograms (kg) |
| Speed | v | metres per second (m/s) |
Exam Tip: Notice that speed is squared in the kinetic energy equation. This means that if you double the speed of an object, its kinetic energy increases by a factor of four (2^2 = 4). This is a very common exam question and is also important for understanding braking distances.
Example 1: A car of mass 1200 kg is travelling at 15 m/s. Calculate its kinetic energy.
E_k = 0.5 x m x v^2 E_k = 0.5 x 1200 x 15^2 E_k = 0.5 x 1200 x 225 E_k = 135 000 J (or 135 kJ)
Example 2: A tennis ball of mass 0.058 kg is hit at a speed of 50 m/s. Calculate its kinetic energy.
E_k = 0.5 x m x v^2 E_k = 0.5 x 0.058 x 50^2 E_k = 0.5 x 0.058 x 2500 E_k = 72.5 J
Example 3: A cyclist has 4000 J of kinetic energy and a mass of 80 kg. Calculate the speed of the cyclist.
E_k = 0.5 x m x v^2 4000 = 0.5 x 80 x v^2 4000 = 40 x v^2 v^2 = 100 v = 10 m/s
Exam Tip: When rearranging the kinetic energy equation to find speed, remember to take the square root as the final step. Write out each step clearly to gain full method marks even if your final answer is wrong.
Gravitational potential energy (GPE) is the energy stored in an object because of its position in a gravitational field — specifically, its height above a reference point (usually the ground).
The higher an object is lifted, and the more massive it is, the more gravitational potential energy it has.
E_p = m x g x h
Where:
| Quantity | Symbol | Unit |
|---|---|---|
| Gravitational potential energy | E_p | joules (J) |
| Mass | m | kilograms (kg) |
| Gravitational field strength | g | newtons per kilogram (N/kg) |
| Height | h | metres (m) |
Exam Tip: The question will tell you whether to use g = 9.8 N/kg or g = 10 N/kg. Always check. If no value is given, use 9.8 N/kg unless the question says "assume g = 10 N/kg."
Example 1: A book of mass 0.5 kg is lifted to a shelf 2 m above the floor. Calculate the gravitational potential energy gained. (Use g = 9.8 N/kg.)
E_p = m x g x h E_p = 0.5 x 9.8 x 2 E_p = 9.8 J
Example 2: A crane lifts a steel beam of mass 500 kg to a height of 12 m. Calculate the GPE stored. (Use g = 9.8 N/kg.)
E_p = m x g x h E_p = 500 x 9.8 x 12 E_p = 58 800 J (or 58.8 kJ)
Example 3: A ball of mass 0.2 kg is thrown upward and gains 7.84 J of gravitational potential energy. How high does it reach? (Use g = 9.8 N/kg.)
E_p = m x g x h 7.84 = 0.2 x 9.8 x h 7.84 = 1.96 x h h = 4 m
When an object falls or rises, energy is transferred between gravitational potential energy and kinetic energy. If we ignore air resistance:
In a closed system (no air resistance or friction), the total of kinetic energy and gravitational potential energy remains constant.
GPE lost = KE gained (when falling) KE lost = GPE gained (when rising)
graph TD
A["Object at maximum height"] -->|"Falls: GPE decreases"| B["Object mid-fall"]
B -->|"KE increases"| C["Object at ground level"]
D["Maximum GPE, zero KE"] --> E["GPE converting to KE"] --> F["Zero GPE, maximum KE"]
style A fill:#aaddff,stroke:#0066cc
style C fill:#ffddaa,stroke:#cc6600
A ball of mass 0.5 kg is dropped from a height of 10 m. Assuming no air resistance and g = 10 N/kg, calculate the speed of the ball just before it hits the ground.
Step 1: Calculate GPE at the top. E_p = m x g x h = 0.5 x 10 x 10 = 50 J
Step 2: By conservation of energy, all GPE converts to KE at the bottom. E_k = 50 J
Step 3: Use the KE equation to find speed. E_k = 0.5 x m x v^2 50 = 0.5 x 0.5 x v^2 50 = 0.25 x v^2 v^2 = 200 v = 14.1 m/s (to 1 decimal place)
Exam Tip: Conservation of energy calculations combining GPE and KE are among the most commonly examined multi-step problems. Always show clearly: (1) the energy at the start, (2) equating the two energy stores, (3) rearranging and solving. This gains full marks.
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