Answer :
Final answer:
The spring coefficient (spring constant, k) of a spring that stores 4000 joules of energy when stretched 0.6 meters is approximately 22222.22 N/m. This calculation uses the spring energy formula with the rearranged equation to solve for k.
Explanation:
The question is asking to find the spring coefficient (also known as the spring constant, k) of a spring that stores 4000 joules of energy when it is stretched by 0.6 meters. To solve this, we can use the spring energy formula: E = 1/2 × k × x², where E is the energy stored in the spring in joules, k is the spring constant in Newtons per meter (N/m), and x is the displacement of the spring in meters. Rearranging the formula to solve for k gives us k = 2E / x².
Substituting the given values, k = 2 × 4000 J / (0.6 m)², we calculate the spring constant to be approximately 22222.22 N/m. This means that the spring is quite stiff, requiring a significant amount of force to stretch it by each meter.
Answer:
222.2N/m
Explanation:
Given parameters:
Elastic potential energy = 4000J
Extension = 6m
Unknown:
Spring coefficient = ?
Solution:
The elastic potential energy is the energy stored within a string.
It is expressed as;
EPE = [tex]\frac{1}{2}[/tex] k e²
k is the spring constant
e is the extension
4000 = [tex]\frac{1}{2}[/tex] x k x 6²
8000 = 36k
k = 222.2N/m
