A spiral spring 20cm long is stretched to 25cm by a load of 50N.What will be its length when streched by 100N assumimg that the elastic limit is not reached?
k = 50N/(25-20)cm = 10N/cm.
L = 20 + 100N*1cm/10N = 20 + 10 = 30 cm.
0.05
To solve this problem, we can use Hooke's Law, which states that the force applied to an elastic material is directly proportional to the extension or compression of that material. The equation can be written as:
F = k * Δx
Where:
F is the force applied (in Newtons),
k is the spring constant (in N/m), and
Δx is the change in length (in meters).
In this case, we are given the initial length of the spring (20 cm = 0.2 m), the force applied (50 N), and the change in length (25 cm - 20 cm = 5 cm = 0.05 m).
First, we can calculate the spring constant (k) using the initial length and the force applied:
k = F / Δx
k = 50 N / 0.05 m
k = 1000 N/m
Now that we know the spring constant, we can calculate the change in length (Δx) when a force of 100 N is applied:
Δx = F / k
Δx = 100 N / 1000 N/m
Δx = 0.1 m
Finally, we can calculate the total length of the spring when stretched by 100 N:
Length = initial length + Δx
Length = 0.2 m + 0.1 m
Length = 0.3 m
Therefore, the spring will be stretched to a length of 0.3 meters when stretched by 100 Newtons, assuming the elastic limit is not reached.