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?
If just a force of 4.6N, then it extends
4.6/6.2 * 0.034 = ____ cm
If you mean an additional 4.6N, then it extends an additional
(6.2+4.6)/6.2 = ____ cm
Now just add those to the original 9.2 cm length
To find the length of the spiral spring when a force of 4.6 N is applied, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the extension or compression of the spring. The formula for Hooke's Law is:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant, and
x is the extension or compression of the spring.
To find the spring constant (k), we can rearrange the formula:
k = F / x
Given that a force of 6.2 N extended the spring by 0.034 m, we can calculate the spring constant:
k = 6.2 N / 0.034 m = 182.35 N/m
Now, we can use the same formula to find the new length of the spring when a force of 4.6 N is applied. Rearranging the formula:
x = F / k
Substituting the values:
x = 4.6 N / 182.35 N/m ≈ 0.0252 m
Finally, to find the length of the spring, we add the extension to the original length:
Length = Original Length + Extension
Length = 9.2 cm + 0.0252 m = 0.092 m + 0.0252 m = 0.1172 m
Therefore, the length of the spiral spring when a force of 4.6 N is applied, without exceeding the elastic limit, is approximately 0.1172 meters.