A horizontal spring, resting on a frictionless tabletop, is stretched 14 from its unstretched configuration and a 1.00- mass is attached to it. The system is released from rest. A fraction of a second later, the spring finds itself compressed 3.6 from its unstretched configuration.

How does its final potential energy compare to its initial potential energy? (Give your answer as a ratio, final to initial.)

You need stretch and mass dimensions after your numbers. However, in this case we can answer it anyway.

Potential Energy of a spring is proportional to the square of the stretch (or compression) dimension. 3.6 is 25.7% of 14. The P.E. will be (0.257)^2 = 6.6% of the original maximum-stretch value. The rest (93.4%) of the original energy becomes kinetic energy at that point

To determine how the final potential energy compares to the initial potential energy, we first need to calculate the potential energy at each position.

The formula for potential energy in a spring is given by:

PE = (1/2)kx^2

where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the spring is initially stretched by 14 cm, which is equivalent to 0.14 meters. The spring constant, denoted by k, is not given, but we can calculate it using Hooke's Law:

F = -kx,

where F is the force exerted by the spring and x is the displacement from the equilibrium position. In this case, the mass of the object attached to the spring is 1 kg, and the displacement is 0.14 meters. Therefore:

F = (1 kg)(9.8 m/s^2) = 9.8 N.

Since the force exerted by the spring is equal to the weight of the object attached to it (assuming negligible mass of the spring), we can substitute this force back into Hooke's Law:

9.8 N = -k(0.14 m).

Solving for k:

k = -9.8 N / (0.14 m) = -70 N/m.

Now that we have the spring constant, we can calculate the initial potential energy:

PE_initial = (1/2)(-70 N/m)(0.14 m)^2
= (1/2)(-70 N/m)(0.0196 m^2)
= -0.686 Joules.

Note that the negative sign indicates that the potential energy is in the compressed state.

Next, we need to calculate the potential energy when the spring is compressed by 3.6 cm, which is equivalent to 0.036 meters:

PE_final = (1/2)(-70 N/m)(0.036 m)^2
= (1/2)(-70 N/m)(0.001296 m^2)
= -0.04536 Joules.

Again, the negative sign indicates the compressed state.

Finally, we can determine the ratio of the final potential energy to the initial potential energy:

PE_final / PE_initial = -0.04536 J / -0.686 J
≈ 0.066.

Therefore, the final potential energy is approximately 0.066 times the initial potential energy, or in a ratio, final to initial, it is 0.066:1.