Hooke's Law asserts that the force required to either stretch or compress a spring from its natural length varies as the displacement. A certain spring is stretched 12 inches beyond its natural length by a 3 lb. weight. Express the force in terms of the displacement for this spring.
force in term of dispacement?
3lb/12in or 1lb/4inches
Oh really? Is that all was meant? Perhaps I was thinking along too complex lines.
Thanks, I appreciate it.
Probably 4oz/inch iz best.
To express the force in terms of the displacement for a spring, we can use Hooke's Law, which states that the force is directly proportional to the displacement. Hooke's Law can be written as:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant (a measure of the stiffness of the spring),
x is the displacement from the spring's natural length.
To find the force in this specific scenario, we are given that the spring is stretched 12 inches beyond its natural length by a 3 lb. weight. The displacement (x) is therefore 12 inches.
Now, we need to find the spring constant (k) to calculate the force. The spring constant is a characteristic property of the spring and varies for different springs. It defines how stiff the spring is.
To find the spring constant, we can rearrange Hooke's Law:
k = F / x
We are given that the weight applied is 3 lb. and the displacement is 12 inches. Plugging these values into the equation:
k = 3 lb. / 12 inches
Simplifying:
k = 0.25 lb./inch
Now that we have the spring constant (k), we can calculate the force (F) using Hooke's Law:
F = k * x = 0.25 lb./inch * 12 inches
F = 3 lb.
Therefore, the force required to stretch the spring by 12 inches is 3 lb.