An 80N-force extends a spring of natural length 8m by 0.4m what will be the length of the spring when the applied force is 100N?
To solve this problem, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension or compression of the spring. Mathematically, we can represent this as F = kx, where F is the force applied, k is the spring constant, and x is the extension or compression of the spring.
We have been given that a force of 80N extends the spring by 0.4m. We can use this information to find the spring constant (k):
80N = k * 0.4m
k = 80N / 0.4m
k = 200 N/m
Now we can use the spring constant to find the length of the spring when a force of 100N is applied. Let's assume the new length of the spring is L:
100N = 200 N/m * (L - 8m)
100N = 200 N/m * L - 1600 N/m
200 N/m * L = 100N + 1600 N/m
200 N/m * L = 1700 N/m
L = 1700 N/m / 200 N/m
L = 8.5m
Therefore, when a force of 100N is applied, the length of the spring will be 8.5m.