A Spring (k=790 N/m) has a length of 48 cm when zero net force is applied to it. What will its length be when 230 N of force is applied to stretch it?
this one is messed up
To find the new length of the spring when a force of 230 N is applied to stretch it, we can use Hooke's Law, which states that the extension of the spring is directly proportional to the applied force.
Hooke's Law can be written as:
F = k * Δx
Where F is the applied force, k is the spring constant (stiffness), and Δx is the extension of the spring.
In this case, we are given the spring constant (k = 790 N/m) and the applied force (F = 230 N). We need to find the extension of the spring (Δx) and then calculate the new length of the spring.
Rearranging the equation, we have:
Δx = F / k
Substituting the given values, we get:
Δx = 230 N / 790 N/m
Δx ≈ 0.291 M
The extension of the spring is approximately 0.291 meters.
To find the new length of the spring, we need to add the extension (Δx) to its original length. The original length of the spring is given as 48 cm, which is equivalent to 0.48 meters.
New length = original length + extension
New length = 0.48 m + 0.291 m
New length ≈ 0.771 m
Therefore, when a force of 230 N is applied to stretch the spring, its length will be approximately 0.771 meters.
Divide the force (230 N) but the spring constant k (790 N/m), and you will get the increase in length. You should get 0.3026 m, which is 30.26 cm.
Add that to the original length to get the new spring length with that force applied. m. Assume that the spring is being stretched. (It could also be compressed, in which case you would subtract the 30.36 m)
230 divided by 790
= .29113924m
= .291m