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.