Practise calculations and explanations involving kinetic, gravitational potential and elastic potential energy. Try each question before revealing the worked answer. These are original AQA-style practice questions, not past-paper questions.
Calculation reminder: Show your equation, substitutions, working and final unit. Use the value of g given in each question.
Question 1
State the unit used to measure energy. (1 mark)
Reveal answer and marking guidance
Mark value: 1
AO focus: AO1: Knowledge of units.
Bullet-point answer points:
- Joule (J).
Full model answer: Energy is measured in joules (J).
Common pitfalls and how to avoid them: Do not give watts (W): watts measure power, not energy.
Question 2
A 3.0 kg box is lifted vertically by 2.0 m. The gravitational field strength is 10 N/kg.
Calculate the increase in the box’s gravitational potential energy. (2 marks)
Reveal answer and marking guidance
Mark value: 2
AO focus: AO2: Applying an equation.
Bullet-point answer points:
- Ep = mgh (or a correct substitution).
- 3.0 × 10 × 2.0 = 60 J.
Full model answer: Ep = mgh = 3.0 × 10 × 2.0 = 60 J.
Common pitfalls and how to avoid them: Use mass in kg, not weight in N. Use the stated value of g, not a different value.
Question 3
A 5.0 kg trolley travels at 4.0 m/s.
Calculate the trolley’s kinetic energy and state how it changes if its speed doubles. (3 marks)
Reveal answer and marking guidance
Mark value: 3
AO focus: AO2: Calculating and applying squared relationships.
Bullet-point answer points:
- Ek = ½mv2.
- Ek = ½ × 5.0 × 4.02 = 40 J.
- Doubling speed makes energy four times greater: 160 J.
Full model answer: Ek = ½mv2 = ½ × 5.0 × 16 = 40 J. At twice the speed, kinetic energy is four times as large, so it becomes 160 J.
Common pitfalls and how to avoid them: Square the speed, not the whole expression. A doubled speed does not just double kinetic energy.
Question 4
A spring has a spring constant of 250 N/m. It is stretched by 8.0 cm and stays within its limit of proportionality.
Calculate the elastic potential energy stored by the spring. (4 marks)
Reveal answer and marking guidance
Mark value: 4
AO focus: AO2: Unit conversion and multi-step calculation.
Bullet-point answer points:
- Convert 8.0 cm to 0.080 m.
- Select Ee = ½ke2.
- Substitute ½ × 250 × (0.080)2.
- Calculate 0.80 J.
Full model answer: First, e = 8.0 ÷ 100 = 0.080 m. Ee = ½ke2 = ½ × 250 × (0.080)2 = 0.80 J.
Common pitfalls and how to avoid them: Do not enter 8 into an equation requiring metres. Square 0.080 before multiplying; use 250 N/m as k, not as force.
Question 5
A moving trolley has 72 J of kinetic energy and a mass of 4.0 kg.
Calculate the trolley’s speed. Show how you rearrange the equation. (5 marks)
Reveal answer and marking guidance
Mark value: 5
AO focus: AO2: Rearrangement and calculation.
Bullet-point answer points:
- Start with Ek = ½mv2.
- Rearrange to v2 = 2Ek / m.
- Substitute v2 = (2 × 72) / 4.0.
- Evaluate v2 = 36.
- Take the square root: v = 6.0 m/s.
Full model answer: Ek = ½mv2, so 2Ek = mv2 and v2 = 2Ek/m = (2 × 72)/4.0 = 36. Therefore v = √36 = 6.0 m/s.
Common pitfalls and how to avoid them: A square root is essential when finding speed; 36 is v², not the final speed. State m/s rather than joules.
Question 6
A 2.0 kg cart starts at rest at the top of a 5.0 m high track. The gravitational field strength is 10 N/kg. Assume no energy transfers to the surroundings in the ideal case.
Explain the energy transfer as the cart descends, calculate its theoretical speed at the bottom, and explain why its actual speed may be lower. (6 marks)
Reveal answer and marking guidance
Mark value: 6
AO focus: AO1: Energy-store explanation; AO2: Linked calculations; AO3: Explaining a real-world difference.
Bullet-point answer points:
- The gravitational potential energy store decreases as height decreases.
- Calculate initial energy: Ep = 2.0 × 10 × 5.0 = 100 J.
- In the ideal case 100 J transfers to the kinetic energy store.
- Use ½mv2 = 100 J; hence ½ × 2.0 × v2 = 100.
- Solve to get v = 10 m/s.
- In reality friction and air resistance transfer some energy to thermal stores (and sound), so the kinetic store is smaller and speed is lower.
Full model answer: As the cart moves down the track, its gravitational potential energy store decreases and its kinetic energy store increases. Initially, Ep = mgh = 2.0 × 10 × 5.0 = 100 J. With no energy transferred to the surroundings, all 100 J becomes kinetic energy. Since Ek = ½mv2, 100 = ½ × 2.0 × v2, so v2 = 100 and the theoretical speed is 10 m/s. In a real experiment, friction and air resistance transfer some energy to thermal stores and sound. Less energy remains in the kinetic store, so the cart travels more slowly.
Common pitfalls and how to avoid them: Do not say energy is “lost” or “used up”: it is transferred to other stores. Do not use Ep = mgh directly to calculate speed without connecting the two equations.