A 10.-newton force is required to hold a

stretched spring 0.20 meter from its rest position. What is the potential energy stored in the
stretched spring?

It's actually 1J. You forgot to divide by one! Thank you!

It's definitly 1J

Double check that. Are you sure that it is 1J?

Guys it is 1 J

To find the potential energy stored in a stretched spring, you can use the formula:

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

Where:
- k is the spring constant (a measure of the stiffness of the spring),
- x is the displacement from the rest position.

In this case, the force (F) required to hold the spring at a specific displacement could be used to find the spring constant (k). The formula is:

F = k * x

Rearranging the equation, we have:

k = F / x

Now we can substitute the given values:

F = 10 N (given)
x = 0.20 m (given)

k = 10 N / 0.20 m = 50 N/m

Using the spring constant (k) and the displacement (x), we can calculate the potential energy:

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

Potential Energy = (1/2) * 50 N/m * (0.20 m)^2

Potential Energy = (1/2) * 50 N/m * 0.04 m^2

Potential Energy = 1 N*m = 1 Joule

Therefore, the potential energy stored in the stretched spring is 1 Joule.

I believe the answer is 1/2kx^2=Fd=10N(0.2m)=2J. Hopefully someone checks this.