A spring of 40cm long is stretched to load of 50N. What will be it's length when stretched by load of 100N. Assuming that the elastic limit is not reached
To find the length of the spring when stretched by a load of 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 formula: F = kx
Where:
F = force applied to the spring (N)
k = spring constant (N/m)
x = extension of the spring (m)
Given that the spring is 40cm (0.4m) long when stretched by a load of 50N, we can calculate the spring constant (k).
50N = k * 0.4m
k = 50N / 0.4m
k = 125 N/m
Now, we can use this spring constant to determine the length of the spring when stretched by a load of 100N.
100N = 125 N/m * x
x = 100N / 125 N/m
x = 0.8 m
Therefore, the length of the spring when stretched by a load of 100N will be 0.8 meters.
To answer this question, we can use Hooke's law, which states that the extension of a spring is directly proportional to the force applied to it.
Hooke's law can be mathematically represented as F = kx, where F is the force applied, k is the spring constant, and x is the extension or deformation of the spring from its original length.
Let's assume that the spring constant of the given spring is k.
Given that the original length of the spring is 40 cm (or 0.4 m) and the load is 50 N, we can calculate the spring constant using Hooke's law:
50 N = k * 0.4 m
We can solve this equation for k:
k = 50 N / 0.4 m
k = 125 N/m
Now that we have found the spring constant, we can use Hooke's law again to find the length of the spring when a load of 100 N is applied.
100 N = 125 N/m * x
Let's solve this equation for x:
x = 100 N / 125 N/m
x = 0.8 m
Therefore, the spring will be stretched to a length of 0.8 meters (or 80 cm) when a load of 100N is applied, assuming the elastic limit is not reached.