A spring 30cm long is stretched to 35cm by a load of 50N.What will be its length when stretched by 100N assuming that the elastic limit is not reached?
We can use Hooke's Law to solve this problem. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or change in length of the spring. Mathematically, it can be represented as:
F = kx
Where:
F = force applied to the spring (in Newtons)
k = spring constant (in Newtons per meter)
x = change in length of the spring (in meters)
We are given that the spring is stretched from 30cm (0.3m) to 35cm (0.35m) by a load of 50N. Let's use these values to find the spring constant (k):
50N = k(0.35m - 0.3m)
50N = k(0.05m)
Now, solve for k:
k = 50N / 0.05m
k = 1000 N/m
Now, we can use the spring constant to find the change in length (x) when a force of 100N is applied:
100N = (1000 N/m) x
x = 100N / 1000 N/m
x = 0.1m
Finally, we can calculate the final length of the spring when stretched by 100N:
New length = original length + change in length
New length = 0.35m + 0.1m
New length = 0.45m
Therefore, the spring will be 45cm long when stretched by 100N.