A spring 20cm long is stretched to 25cm by a load of 50n, what will be its length when stretched by 100n , assuming that elastic limit is not exceeded?
if 50N expands it 5cm, 100N will expand it 10 com, or to 30cm length.
25 - 20 = 5 cm,
50/5 = 100/X,
L = 20 + x.
To find the length of the spring when stretched by 100N, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to its extension.
Hooke's Law equation: F = k * x
Where:
F = Force applied to the spring (in Newtons)
k = Spring constant (in Newtons per meter)
x = Extension of the spring (in meters)
First, we need to find the spring constant (k) using the given information. We can rearrange Hooke's Law equation to solve for k:
k = F / x
Now, let's calculate the spring constant:
Given force (F1) = 50N
Initial extension (x1) = 25cm = 0.25m
k = F1 / x1
k = 50N / 0.25m
k = 200 N/m
Now that we have the spring constant, we can find the extension (x2) when the load (F2) is 100N.
Given force (F2) = 100N
Using Hooke's Law, we can rearrange the equation to solve for the extension:
x2 = F2 / k
Substituting the values into the equation:
x2 = 100N / 200 N/m
x2 = 0.5m
Therefore, the length of the spring when stretched by 100N is 0.5 meters or 50cm.