There 46.5J of elastic potential energy stored in a spring. How far is the spring stretched if the spring constant is 455N/m

PE = (1/2)kx^2

(46.5) = (1/2)(455)(x)^2
x = 0.452 meters

To find the distance the spring is stretched, we will use the formula for the potential energy of a spring:

Elastic Potential Energy (PE) = 1/2 * k * x^2

Where:
PE is the elastic potential energy stored in the spring
k is the spring constant
x is the distance the spring is stretched

In this case, we know the elastic potential energy (PE) is 46.5J and the spring constant (k) is 455N/m. We need to solve for x.

Rearranging the formula, we have:

x^2 = (2 * PE) / k

x^2 = (2 * 46.5J) / 455N/m

x^2 = 0.204

Taking the square root, we find:

x ≈ 0.452 meters (rounded to three decimal places)

Therefore, the spring is stretched approximately 0.452 meters.