A spring of natural length 3m is extended by 0.01m by a force of 4N. What will never it's length when the applied force is 12N?
To solve this problem, we need to understand Hooke's law, which states that the force required to extend or compress a spring is directly proportional to the displacement from its natural length.
Mathematically, Hooke's law is given by the equation:
F = k * x
Where:
- F is the force applied to the spring,
- k is the spring constant, which measures the stiffness of the spring,
- x is the displacement from the natural length of the spring.
In this case, we are given that the natural length of the spring (L0) is 3m, and it is extended by 0.01m with a force of 4N. We can use this information to find the spring constant (k).
First, we rearrange Hooke's law equation to solve for k:
k = F / x
k = 4N / 0.01m
k = 400 N/m
Now that we have the spring constant, we can determine the length (L) of the spring when a force of 12N is applied. Again, we can use Hooke's law:
F = k * x
12N = 400 N/m * x
Now, let's rearrange the equation to solve for x:
x = F / k
x = 12N / 400 N/m
x = 0.03m
Therefore, when a force of 12N is applied, the spring will extend by 0.03m from its natural length of 3m. Its total length will be:
L = L0 + x
L = 3m + 0.03m
L = 3.03m