Question

bo pulls a slinky, normally 0.05m long, to a length of 2m. if the spring constant of a slinky is 10N/m, what is the elastic potential energy stored in the stretched slinky

Answer 1

The stretched slinky has an elastic potential energy of 19.0125 Joules, calculated using the formula ½ k x², where k is the spring constant, and x is the displacement.

Step-by-step explanation:

Elastic Potential Energy Calculation

To calculate the elastic potential energy stored in a stretched slinky, we use the formula for spring potential energy, which is derived from Hooke's Law:

Potential Energy (PEs) = ½ k x²

Where:

k is the spring constant (10 N/m)

x is the displacement from the equilibrium position (2 m - 0.05 m)

Substitute the given values into the equation:

PEs = ½ (10 N/m) (2 m - 0.05 m)²

First, calculate the displacement (x):

x = 2 m - 0.05 m = 1.95 m

Then, substitute x into the equation to find the elastic potential energy:
PEs = ½ (10 N/m) (1.95 m)²
PEs = ½ (10 N/m) (3.8025 m²)
PEs = (5 N/m) (3.8025 m²)
PEs = 19.0125 J

The stretched slinky has an elastic potential energy of 19.0125 Joules (J).