A force of 2N extends a spring of natural length of 0.01m, what will be the length of the applied force is 10N
Five times the force results in five times the extension.
F = k X
so
X = F / k
To find the length of the spring when a force of 10N is applied, we can use Hooke's Law which states that the force applied to a spring is directly proportional to the extension or compression of the spring.
Mathematically, Hooke's Law can be represented as:
F = k * x
Where:
F = Force applied to the spring
k = Spring constant
x = Extension or compression of the spring
Since we know that a force of 2N extends the spring by 0.01m, we can calculate the spring constant (k) by rearranging Hooke's Law:
k = F / x
k = 2N / 0.01m
k = 200 N/m
Now, we can use the spring constant to find the length of the spring when a force of 10N is applied:
F = k * x
10N = 200 N/m * x
x = 10N / 200 N/m
x = 0.05m
Therefore, the length of the spring when a force of 10N is applied is 0.05 meters.
To calculate the length of an extended spring, we can use Hooke's Law. Hooke's Law states that the force applied to extend or compress a spring is directly proportional to the displacement of the spring from its natural length.
The formula for Hooke's Law is:
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 displacement of the spring from its natural length.
In this case, we have a force of 2N that extends the spring by 0.01m. We need to find the length of the spring when a force of 10N is applied.
First, we can calculate the spring constant (k) using the initial force and displacement.
F = k * x
2N = k * 0.01m
k = 2N / 0.01m
k = 200 N/m
Now that we have the spring constant (k), we can use it to find the displacement (x) when a force of 10N is applied.
F = k * x
10N = 200 N/m * x
x = 10N / 200 N/m
x = 0.05m
Therefore, when a force of 10N is applied, the spring will be extended by 0.05m.