a spring with spring constant k= 100N/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 6cm longer than it was. 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 2cm shorter than its equilibrium length. How much elastic potential energy is stored in this spring?
I am not certain of what you are stuck on. At equilibrium, there is no PE stored in the spring. When using F=kx, and KE=1/2 k x^2, be certain x is in meters.
a) Oh spring, you're so potential! The elastic potential energy stored in a spring at its equilibrium length can be calculated using the formula: E = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium. Since the spring is at its equilibrium length, x is zero. Therefore, the elastic potential energy is also zero. We've got a relaxed and stress-free spring here!
b) Ah, the spring is getting a stretch, huh? Now it's 6cm longer than before. Let's find out how much elastic potential energy it's storing. To do that, we'll need to calculate the new displacement from the equilibrium length, which is 6cm or 0.06m. Plug that into the formula: E = (1/2)kx^2. With k = 100N/m and x = 0.06m, we get E = (1/2)(100)(0.06)^2 = 0.18 J. The spring is feeling energetic with 0.18 Joules of elastic potential energy!
c) Ah, a different spring with a new spring constant, and it's being compressed this time! The spring constant is four times larger than in part (a), so let's call it k' = 4k = 400N/m. The new displacement from equilibrium is -2cm or -0.02m. Applying the formula, E = (1/2)k'x^2, we get E = (1/2)(400)(-0.02)^2 = 0.08 J. The compressed spring is still storing some potential, with 0.08 Joules of elastic potential energy!
a) To calculate the elastic potential energy stored in a spring, you can use the formula:
Elastic Potential Energy (E) = 0.5 * 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 0.
Therefore, the elastic potential energy is:
E = 0.5 * 100 * 0^2 = 0 Joules
b) When the spring is stretched by 6 cm, the displacement (x) is 6 cm or 0.06 m.
Using the same formula, the elastic potential energy will be:
E = 0.5 * 100 * (0.06)^2 = 0.18 Joules
c) Let's assume the spring constant for the second spring is k'. Given that the spring constant is four times larger than in part (a), we have k' = 4 * 100 = 400 N/m.
When the spring is compressed by 2 cm, the displacement (x) is -2 cm or -0.02 m (negative because it is compressed).
Using the same formula, the elastic potential energy will be:
E = 0.5 * 400 * (-0.02)^2 = 0.08 Joules
To answer these questions, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium. The formula for elastic potential energy stored in a spring is given by U = (1/2)kx^2, where U is the elastic potential energy, k is the spring constant, and x is the displacement from equilibrium.
a) Since the spring is at its equilibrium length, the displacement (x) is zero. Therefore, the elastic potential energy stored in the spring is zero.
b) To calculate the elastic potential energy with the spring stretched by 6 cm, we need to find the new displacement (x). Given that the spring is stretched by 6 cm, the displacement from equilibrium is x = 0.06 m (since 1 cm is equal to 0.01 m). Now, we can use the formula U = (1/2)kx^2 to find the elastic potential energy. Plugging in the values, we get U = (1/2) * 100 N/m * (0.06 m)^2 = 0.18 J.
c) In this case, we have a different spring with a spring constant four times larger than in part (a). Let's assume the spring constant is now 4 * 100 N/m = 400 N/m. The spring is compressed by 2 cm, which gives us a displacement of x = -0.02 m (negative because it is compressed). We can calculate the elastic potential energy using the formula U = (1/2)kx^2. Plugging in the values, we get U = (1/2) * 400 N/m * (-0.02 m)^2 = 0.002 J.
Therefore, the answers to the questions are:
a) 0 J
b) 0.18 J
c) 0.002 J