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.