A spring 40cm long is stretched to 45cm by a load of 50N . What will be its length when stretched by 100N , assuming that the elastic limited is not reached
you don't have to calculate the question -- you need to calculate the answer.
and why ask twice?
100/(x-40) = 50/(45-40)
To solve this problem, we can use Hooke's Law, which states that the extension of an elastic material is directly proportional to the applied force, as long as the material is within its elastic limit.
Hooke's Law can be expressed as:
F = k * x
Where:
F = Force applied (in Newtons)
k = Spring constant (in Newtons per meter)
x = Extension or compression of the spring (in meters)
We can use this formula to find the spring constant (k) for the given problem:
F1 = k * x1 ----(1)
F2 = k * x2 ----(2)
Where:
F1 = 50 N (load applied)
x1 = 45 cm (extension)
Plugging in these values, we can solve for the spring constant:
50 N = k * 0.45 m
k = 50 N / 0.45 m
k = 111.11 N/m (approximated to two decimal places)
Now that we know the spring constant, we can use it to find the extension when a load of 100N is applied:
100 N = 111.11 N/m * x2
Solving for x2, we get:
x2 = 100 N / 111.11 N/m
x2 ≈ 0.90 m
Finally, we can convert the extension from meters to centimeters:
x2 = 0.90 m * 100 cm/m
x2 = 90 cm
Therefore, when stretched by 100N, the spring will be approximately 90cm long.