A force of 5n applied to a felastic spring 2m makes it to extendly 0.01 what will be the new length if force is 10n is applied
F = kx, so
x/10 = .01/2
where x is the amount of extension.
delta x is the amount of extension
5 = k (.01)
k = 5/.01 = 500 N/m
10 = 500 (delta X)
delta x = 1/50 = .02 meter
so new length = 2,02 meters
To determine the new length of the elastic spring when a 10N force is applied, we need to understand Hooke's Law, which describes the relationship between the force applied to a spring and the extension it experiences.
Hooke's Law is expressed as:
F = k * x
where:
F is the force applied to the spring,
k is the spring constant,
x is the extension or compression of the spring.
In this case, we are given:
Force 1 (F1) = 5N
Extension 1 (x1) = 0.01m
Force 2 (F2) = 10N
Extension 2 (x2) = ?
The first step is to find the spring constant (k). It can be calculated using the Hooke's Law formula rearranged as:
k = F1 / x1
k = 5N / 0.01m
k = 500 N/m
Now, we can use the calculated spring constant to determine the new extension (x2) using the rearranged Hooke's Law formula:
x2 = F2 / k
x2 = 10N / 500 N/m
x2 = 0.02m
Therefore, the new length of the elastic spring when a 10N force is applied is 0.02 meters.