A sping of natural length 3m is extended by 0.01m by a force of 4m,what will be its length when the applied force is 12n
F=kx, so 3 times the force means 3 times the extension
To answer this question, we need to apply Hooke's Law, which states that the force needed to stretch or compress a spring is directly proportional to the displacement of the spring from its natural length. Mathematically, this can be written as:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant (a measure of its stiffness),
x is the displacement of the spring from its natural length.
Given that the spring's natural length is 3m and it is extended by 0.01m with a force of 4N, we can determine the spring constant:
4N = k * 0.01m
Rearranging the equation to solve for k:
k = 4N / 0.01m
k = 400 N/m
Now, we can use this spring constant to calculate the new length of the spring when a force of 12N is applied:
12N = 400 N/m * x
Solving for x:
x = 12N / 400 N/m
x = 0.03m
To find the new length, we add the displacement to the natural length:
New length = 3m + 0.03m
New length = 3.03m
Therefore, the length of the spring when a force of 12N is applied will be 3.03m.
To find the new length of the spring when the applied force is 12N, we can use Hooke's Law, which states that the force exerted on a spring is directly proportional to the extension of the spring.
First, let's find the spring constant (k):
k = F / x
k = 4N / 0.01m
k = 400 N/m
Now, we can use the formula to find the extension of the spring when the applied force is 12N:
x = F / k
x = 12N / 400 N/m
x = 0.03m
To get the new length of the spring, we add the extension to the natural length:
New length = Natural length + Extension
New length = 3m + 0.03m
New length = 3.03m
Therefore, when the applied force is 12N, the length of the spring will be 3.03 meters.