Question 1: Elastic potential energy is the stored energy stored in an object when it is deformed elastically. It is dependent on the amount of deformation and the properties of the material. This energy can be released when the object returns to its original shape.

Answer: Elastic potential energy is the stored energy in an object due to elastic deformation, dependent on the extension or compression and the material's properties.

Question 1: What is the formula for calculating elastic potential energy stored in a stretched or compressed object?

Answer: E = (1/2) k x^2, where E is elastic potential energy, k is the spring constant, and x is the extension or compression.

Question 2: Explain how the elastic potential energy changes when the extension of a spring is doubled.

Answer: The elastic potential energy increases by a factor of four, since it is proportional to the square of the extension (x^2).

Question 1: A spring with a spring constant of 150 N/m is compressed by 0.02 m. Calculate the elastic potential energy stored in the spring. (Use E = (1/2) k x^2)

Answer: E = 0.5 × 150 N/m × (0.02 m)^2 = 0.5 × 150 × 0.0004 = 0.03 Joules

Question 2: A rubber band has a spring constant of 10 N/m. If it is stretched by 0.05 m, what is the elastic potential energy stored?

Answer: E = 0.5 × 10 N/m × (0.05 m)^2 = 0.5 × 10 × 0.0025 = 0.0125 Joules

Question 1: Describe a simple experiment to measure the elastic potential energy stored in a stretched spring. Include the variables that need to be controlled.

Answer: One experiment involves attaching a spring to a fixed point and measuring its extension when a known mass is hung from it. The spring constant can be determined from Hooke's Law (F = kx). The elastic potential energy is then calculated using E = (1/2) k x^2. Variables to control include the mass used, the temperature (to prevent material changes), and ensuring the spring is not overstretched beyond its elastic limit to avoid permanent deformation.

Question 2: List two safety precautions to consider when performing experiments involving elastic deformation of materials.

Answer: Ensure the materials are not overstretched beyond their elastic limit to prevent breakage, and wear safety goggles to protect eyes from potential snapping or sudden release of energy.

Question 1: A metal wire is stretched by applying a force of 50 N. The wire has a length of 2 meters and a spring constant of 2500 N/m. The wire is stretched by 0.02 meters. Calculate the elastic potential energy stored in the wire.

Answer: E = (1/2) × 2500 N/m × (0.02 m)^2 = 0.5 × 2500 × 0.0004 = 0.5 Joules

Question 2: Discuss how the elastic potential energy stored in a material would change if the material's spring constant is increased, assuming the same extension.

Answer: The elastic potential energy would increase proportionally, since E = (1/2) k x^2. Increasing k results in more energy stored for the same extension.

Question 1: Explain how elastic potential energy is utilised in the design of sports equipment such as trampolines or tennis rackets.

Answer: In trampolines and tennis rackets, elastic materials are deformed when force is applied. The elastic potential energy stored during deformation is released suddenly, propelling the user or ball with increased speed and height. Efficient energy transfer relies on the elastic properties of the materials, allowing for maximum energy storage and release without permanent deformation.

Question 2: Identify a technology or industry where understanding elastic potential energy is crucial, and explain its significance.

Answer: The automotive industry relies on elastic potential energy in shock absorbers, where springs absorb and release energy to smooth out road shocks. Understanding this energy helps in designing safer, more comfortable vehicles.