A spring has a spring constant, k, of 441 N/m. How much must this spring be stretched to store 22 J of potential energy?

yes

To answer this question, we need to make use of the equation for the potential energy stored in a spring, which is given by:

Potential Energy (PE) = 0.5 * k * x^2

In this equation:
- PE represents the potential energy stored in the spring
- k is the spring constant (which is given as 441 N/m)
- x is the displacement or stretch of the spring

We are given that the potential energy stored in the spring is 22 J, and we need to find how much the spring must be stretched (x).

So, we can rearrange the formula to solve for x:

x = sqrt(2 * PE / k)

Let's plug in the values now:

x = sqrt(2 * 22 J / 441 N/m)

x = sqrt(0.1)

x ≈ 0.316 m

Therefore, the spring must be stretched approximately 0.316 meters to store 22 J of potential energy.