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?
To solve this problem, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, as long as the elastic limit is not reached.
Let's first find the spring constant (k) using the given information:
k = F / x,
where k is the spring constant, F is the force applied, and x is the extension of the spring.
Here, the force (F) is 50N and the extension (x) is 35cm - 30cm = 5cm = 0.05m.
k = 50N / 0.05m = 1000 N/m.
Now, let's calculate the length of the spring (L) when stretched by 100N:
L = x + L0,
where L is the length of the spring when stretched, x is the extension of the spring, and L0 is the original length of the spring.
Here, the force (F) is 100N, and we need to find x, L0, and L.
Using Hooke's Law, we can calculate the extension:
F = k * x,
x = F / k,
x = 100N / 1000 N/m = 0.1m.
Now, let's substitute the known values into the equation for L:
L = x + L0,
L = 0.1m + 0.30m = 0.40m.
Therefore, when stretched by 100N, the length of the spring will be 0.40m.