A force of 6.2 N applied to the spiral spring of length 9.2 cm causes it to extend by 0.034 m. What is the length of the spiral spring when a force of 4.6 N is applied to it provided the elastic limit is not exceeded?
F = k x where x is extension
0.034 * (4.6 / 9.2) meters
the ans is not in the option given
To find the length of the spiral spring when a force of 4.6 N is applied to it, we can make use of Hooke's Law, which states that the extension of an elastic material is directly proportional to the force applied to it, as long as the elastic limit is not exceeded.
Hooke's Law can be expressed as:
F = k * x
Where:
- F is the force applied to the spring
- k is the spring constant (a measure of the stiffness of the spring)
- x is the extension or deformation of the spring
To find the spring constant (k), we can rearrange the equation:
k = F / x
Now, we can calculate the spring constant using the force of 6.2 N and the extension of 0.034 m:
k = 6.2 N / 0.034 m
k ≈ 182.35 N/m
Once we have the spring constant, we can use it to find the extension of the spring when a force of 4.6 N is applied. Rearranging Hooke's Law equation again:
x = F / k
Substituting the values:
x = 4.6 N / 182.35 N/m
x ≈ 0.0252 m
Now, to find the length of the spring when this extension occurs, we subtract the extension (0.0252 m) from the original length (9.2 cm = 0.092 m):
Final length = Original length - Extension
Final length = 0.092 m - 0.0252 m
Final length ≈ 0.0668 m or 6.68 cm
Therefore, when a force of 4.6 N is applied to the spring, the length of the spiral spring is approximately 6.68 cm.