Question 1: Elastic potential energy is stored in objects that can be stretched or compressed. When an elastic object is deformed, it stores energy which can be released when it returns to its original shape.

Answer: Elastic potential energy is stored in objects when they are stretched or compressed, due to their deformation.

Question 1: Define elastic potential energy.

Answer: Elastic potential energy is the energy stored in an object when it is stretched or compressed.

Question 2: Explain how the elastic potential energy is related to the deformation of an object.

Answer: Elastic potential energy is proportional to the amount of deformation, such as the extension or compression, of the object. The more an object is deformed, the more energy it stores, up to its elastic limit.

Question 1: A spring with a spring constant of 200 N/m is stretched by 0.05 metres. Calculate the elastic potential energy stored in the spring. Show your working.

Answer: E = 0.5 × 200 N/m × (0.05 m)² = 0.5 × 200 × 0.0025 = 0.25 Joules

Question 2: If a rubber band stores 0.02 Joules of elastic energy when stretched, and the extension is 0.01 metres, what is the approximate spring constant of the rubber band? Show your working.

Answer: k = 2E / x² = 2 × 0.02 / (0.01)² = 0.04 / 0.0001 = 400 N/m

Question 1: Describe a simple experiment to measure the spring constant of a spring. Include the variables you would control and safety precautions.

Answer: To measure the spring constant, hang the spring vertically and attach a mass hanger. Add known masses and record the extension each time. Plot extension against force (mass × gravity). The slope of the line gives the spring constant. Control variables include the temperature and ensure the spring is securely attached. Safety precautions involve handling masses carefully to prevent dropping and injury.

Question 1: A spring is stretched to an extension of 0.02 metres and stores 0.016 Joules of elastic energy. If the extension is increased to 0.04 metres, what is the new energy stored? Assume linear behaviour and show your reasoning.

Answer: Since E ∝ x², (0.04 / 0.02)² = 4. So, new energy = 4 × 0.016 J = 0.064 Joules.

Question 2: A student observes that doubling the extension of a spring quadruples the stored elastic energy. Explain why this behaviour occurs.

Answer: Because elastic energy is proportional to the square of the extension (E ∝ x²), doubling the extension increases the energy by a factor of four.

Question 1: Describe how elastic energy is utilised in a car's suspension system.

Answer: Car suspension systems use springs to absorb shocks. When the car hits a bump, the spring compresses and stores elastic energy, which is then released to smooth the ride.

Question 2: Explain how elastic potential energy is important in the design of sports equipment like tennis rackets or golf clubs.

Answer: Sports equipment like tennis rackets and golf clubs are designed to deform elastically during impact. This deformation stores elastic potential energy, which is then transferred to the ball, increasing its speed and distance.