Explanation:

When a spring is stretched or compressed, it stores potential energy in it. The amount of potential energy stored in a spring is given by the formula:

Potential energy = 1/2 kx^2

Where,
k = spring constant
x = displacement from the equilibrium position

Ratio of potential energy:

In this case, we have two springs of spring constants 1500 N m^-1 and 3000 N m^-1 respectively stretched with the same force. Let us assume that the force applied is F. Therefore, the displacement of each spring will be different due to their different spring constants.

Let x1 and x2 be the displacements of the springs of spring constants 1500 N m^-1 and 3000 N m^-1 respectively.

From Hooke's law, we know that F = kx. Therefore, the force applied on each spring will be:

F1 = 1500 x1
F2 = 3000 x2

Since the force applied is the same, we can equate F1 and F2:

1500 x1 = 3000 x2
x1/x2 = 2

Therefore, the ratio of the displacements of the two springs is 2:1.

Now, using the formula for potential energy, we can find the ratio of potential energy stored in the two springs:

Potential energy of spring 1 = 1/2 x 1500 x x1^2
Potential energy of spring 2 = 1/2 x 3000 x x2^2

Substituting x1/x2 = 2, we get:

Potential energy of spring 1/Potential energy of spring 2 = x1^2/x2^2 = 4/1

Therefore, the ratio of potential energy stored in the two springs is 4:1. The spring with the higher spring constant stores more potential energy for the same force applied due to its higher stiffness.