A spring with spring constant k=100 (N/m) is at its equilibrium length.

a) How much elastic potential energy is stored in the spring?
b) The same spring is stretched so that it is 6.0 cm longer than its equilibrium length. How much elastic potential energy is stored in the spring now?
c) A different spring with spring constant four times larger than in part (a) is compressed so that it is 2.0 cm shorter than its equilibrium length. How much elastic potential energy is stored in this spring?

a) zero

b) (1/2) k X^2, where X = 0.06 m

c) In this case, X is -1/3 of the previous example, and k is 4 times greater. Use the same formula and see what you get.

a) The elastic potential energy stored in a spring can be calculated using the formula:

Elastic potential energy (U) = 1/2 * k * x^2

where k is the spring constant and x is the displacement from the equilibrium position.

In this case, since the spring is at its equilibrium length, the displacement (x) is zero. Therefore, the elastic potential energy is also zero.

b) To calculate the elastic potential energy of the stretched spring, we need to find the new displacement (x) first. It is given that the spring is stretched 6.0 cm longer than its equilibrium length.

Displacement (x) = 0.06 m (1 cm = 0.01 m)

Now we can use the same formula as before:

Elastic potential energy (U) = 1/2 * k * x^2

U = 1/2 * 100 * (0.06)^2
U = 0.18 J

Therefore, the elastic potential energy stored in the spring when it is stretched 6.0 cm longer than its equilibrium length is 0.18 J.

c) The spring constant of the different spring is four times larger than in part (a). So, the spring constant (k) for this spring would be:

k = 4 * 100 (since k in part (a) is given as 100 N/m)

k = 400 N/m

The spring is compressed 2.0 cm shorter than its equilibrium length. So, the displacement (x) would be:

Displacement (x) = -0.02 m (negative sign because it is compressed)

Now we can calculate the elastic potential energy using the formula:

Elastic potential energy (U) = 1/2 * k * x^2

U = 1/2 * 400 * (-0.02)^2
U = 0.08 J

Therefore, the elastic potential energy stored in this spring, when it is compressed 2.0 cm shorter than its equilibrium length, is 0.08 J.

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

Elastic potential energy (U) = (1/2) * k * x^2

Where:
k is the spring constant
x is the displacement from the equilibrium position of the spring

a) In this case, the spring is at its equilibrium length, which means x = 0. Therefore, the elastic potential energy stored in the spring is also zero.

b) Now, the spring is stretched 6.0 cm longer than its equilibrium length. Convert 6.0 cm to meters by dividing by 100: x = 6.0 cm / 100 = 0.06 m

Using the formula, the elastic potential energy in the spring is:
U = (1/2) * 100 * (0.06)^2 = 0.18 J (joules)

c) In this part, we have a spring with a different spring constant that is four times larger than in part (a). Let's call this new spring constant k'.

Let's assume the equilibrium length of the different spring in part (c) is also the same as the initial spring in part (a) since the problem didn't provide a specific value.

Now, the spring is compressed 2.0 cm shorter than its equilibrium length. Convert 2.0 cm to meters: x = 2.0 cm / 100 = 0.02 m

Since the new spring constant is four times larger than the initial spring constant (k), we have k' = 4 * k = 4 * 100 = 400 N/m.

Using the formula, the elastic potential energy in the spring is:
U = (1/2) * 400 * (0.02)^2 = 0.08 J (joules)