Final answer:
The potential energy of a spring with a spring constant of 320 N/m when stretched by 6 cm is 0.288 joules. This is calculated using the formula for elastic potential energy: PE = ½kx².
Step-by-step explanation:
To calculate the potential energy (PE) of a spring stretched by 6 cm, we can use the formula for elastic potential energy, PE = ½kx², where k is the spring constant and x is the displacement from the spring's equilibrium position.
Given that the spring constant k is 320 N/m and the displacement x is 6 cm (which is 0.06 meters, since we need to use SI units), the potential energy stored in the spring can be calculated as follows:
PE = ½(320 N/m)(0.06 m)²
First, square the displacement:
(0.06 m)² = 0.0036 m²
Then, plug this value into the formula:
PE = ½(320 N/m)(0.0036 m²)
PE = 0.5 * 320 * 0.0036
PE = 576 * 0.5
PE = 288 * 0.5 = 0.288 J
So, the potential energy of the spring when stretched by 6 cm is 0.288 joules.