Illustrate a vivid scientific concept. Show a cross-sectional view of a spiral spring with its natural length indicated at 30cm. Detail how it's compressed to 20cm. Use visual indicators to denote the force constant of 20N/m. To demonstrate energy conservation, subtly represent the energy stored in the spring but remember to exclude any specific calculations or text.

A spiral spring of natural length of 30.0cm and a constant force of 20N/m is compressed to 20cm. Calculate the energy stored in the spring.

30 cm - 20 cm = 10 cm = .1 meter

F = k x
20 = k (.1)
so
k = 200

E = (1/2) k x^2
= (1/2)(200)(.01)
= 1 Joule

thanks

Why did the spring go to therapy? Because it was under a lot of tension and needed to unwind!

To calculate the energy stored in the spring, we can use the formula for elastic potential energy:

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

Where k is the spring constant and x is the compression or extension distance from the natural length of the spring.

In this case, the spring constant (k) is 20 N/m and the compression distance (x) is 20 cm.

Converting 20 cm to meters, we get:

x = 20 cm = 0.2 m

Now we can calculate the energy stored in the spring:

E = (1/2)(20 N/m)(0.2 m)^2

E = (1/2)(20)(0.04)

E = 0.4 J

So the energy stored in the spring is 0.4 Joules. Just think of it as a little spring-loaded laugh for your physics problem!

To calculate the energy stored in the spring, we need to use Hooke's Law, which states that the force required to compress or extend a spring is directly proportional to the displacement from its natural length.

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

U = (1/2)kx²

Where U is the potential energy, k is the spring constant, and x is the displacement from the natural length of the spring.

In this case, the natural length of the spring is given as 30.0 cm, and the constant force is 20 N/m. The spring is compressed to 20 cm, which means its displacement (x) is 20 cm - 30 cm = -10 cm = -0.10 m (since the displacement is negative due to compression).

Now we can calculate the energy stored in the spring:

U = (1/2)kx²

U = (1/2)(20 N/m)(-0.10 m)²

U = (1/2)(20 N/m)(0.01 m²)

U = 0.10 J

Therefore, the energy stored in the spring is 0.10 Joules.