A uniform bar of iron is supported by a lon, uniform Hooke's Law spring. The spring is cut in half exactly and the two pieces are used to support the same bar. If the whole spring is stretched by 4 cm in the first bar (that is not cut in half) how much would each spring stretch in the second bar?
if they are in parallel? http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinParallel.html
If they are in series:
http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html
To determine how much each spring would stretch in the second bar, we need to consider how the spring constant changes when it is cut in half.
When the spring is cut in half and the two pieces are used to support the same bar, each piece of the spring will have half the original spring constant. This is because the spring constant depends on the physical properties of the spring, such as its material and dimensions.
Let's denote the original spring constant as k_orig, and let's assume that the spring is ideal and obeys Hooke's Law. When the original spring is stretched by 4 cm (0.04 m), it exerts a force given by F = k_orig * x, where x is the stretching distance.
Now, since the second bar is supported by the two half-springs, each half-spring will experience half of the force compared to the original spring. Therefore, the force exerted by each half-spring will be F/2.
To find out how much each spring will stretch, we can use Hooke's Law again. Let's denote the stretching distance of each half-spring as x_2. We can write:
(F/2) = (k_new * x_2),
where k_new is the new spring constant for each half-spring.
Since we know that k_new = k_orig/2 (due to cutting the spring in half), we can rewrite the equation as:
(F/2) = ((k_orig/2) * x_2).
Now, we substitute the known values: F = k_orig * x.
(k_orig * x)/2 = ((k_orig/2) * x_2).
By canceling the k_orig/2 terms, the equation simplifies to:
x/2 = x_2.
Therefore, each half-spring will stretch the same distance as the original spring, which is 4 cm or 0.04 m.
In conclusion, when the whole spring is stretched by 4 cm in the first bar, each half-spring used to support the second bar will stretch by 4 cm as well.