Question 1: Elastic potential energy is the stored energy in an object when it is stretched or compressed. This energy is released when the object returns to its original shape. It is dependent on the amount of deformation and the material's properties.

Answer: Elastic potential energy is stored energy in a stretched or compressed object, dependent on deformation and material properties.

Question 1: What is the main factor that determines the amount of elastic potential energy stored in a stretched spring?

Answer: The amount of stretch or compression (deformation) of the spring.

Question 2: Explain why a rubber band can store elastic potential energy when stretched.

Answer: A rubber band can store elastic potential energy because when stretched, its molecules are deformed from their equilibrium positions, storing energy that is released when it returns to its original shape.

Question 1: A steel spring with a spring constant of 200 N/m is stretched by 0.05 meters. Calculate the elastic potential energy stored in the spring. (Use the formula: U = 0.5 * k * x^2)

Answer: U = 0.5 * 200 N/m * (0.05 m)^2 = 0.5 * 200 * 0.0025 = 0.25 Joules

Question 2: A bungee cord stretches 1.2 meters when a load of 150 N is attached. If the spring constant of the cord is 125 N/m, what is the elastic potential energy stored in the cord?

Answer: U = 0.5 * 125 N/m * (1.2 m)^2 = 0.5 * 125 * 1.44 = 90 Joules

Question 1: Describe a safe procedure for testing the elastic potential energy of a spring in a classroom setting.

Answer: Secure the spring to a fixed point, gradually stretch it while measuring the extension, and ensure students stand clear of the potential energy release zone. Use appropriate tools and avoid overstretching.

Question 2: Identify two variables that affect the amount of elastic potential energy stored in a stretched object.

Answer: The amount of stretch (deformation) and the material's spring constant.

Question 1: A metal wire is stretched by applying a force of 50 N, resulting in an extension of 0.002 meters. If the wire's spring constant is 25,000 N/m, what is the elastic potential energy stored?

Answer: U = 0.5 * 25,000 N/m * (0.002 m)^2 = 0.5 * 25,000 * 4e-6 = 0.05 Joules

Question 1: Describe one real-world example where elastic potential energy is utilized and explain its importance.

Answer: In shock absorbers of vehicles, elastic potential energy is stored and released to smooth out bumps, providing a comfortable ride.

Question 1: A spring with a spring constant of 300 N/m is compressed by 0.1 meters. Calculate the elastic potential energy stored. Then, explain how increasing the spring constant would affect the energy stored for the same compression.

Answer: Using the formula U = 0.5 * k * x^2, U = 0.5 * 300 N/m * (0.1 m)^2 = 1.5 Joules. Increasing the spring constant increases the energy stored for the same compression because a stiffer spring resists deformation more, storing more energy.