To tackle this problem, we need to analyze the forces acting on the system of discs and the spring. The scenario involves two discs attached to a spring, with one disc resting on a surface and the other disc compressing the spring by a distance X when in equilibrium. Our goal is to determine how much further the spring must be compressed to ensure that, upon removing the compressive force, the spring extends enough to lift the lower disc off the table.
Understanding the Forces at Play
First, let’s break down the forces involved. When the upper disc compresses the spring by a distance X, the spring exerts an upward force equal to the weight of the upper disc plus the weight of the lower disc. The spring force can be calculated using Hooke's Law, which states:
F_s = k * X
Here, F_s is the spring force, k is the spring constant, and X is the compression of the spring. In equilibrium, this spring force balances the gravitational forces acting on the discs.
Calculating the Initial Compression
Let’s denote the mass of each disc as m. The total weight of the two discs is:
W = 2mg
At equilibrium, the spring force equals the total weight:
k * X = 2mg
From this equation, we can express the spring constant in terms of the compression:
k = (2mg) / X
Determining Additional Compression Needed
Now, to lift the lower disc off the table, the spring must exert a force greater than the weight of the lower disc. This means that when the compressive force is removed, the spring must extend enough to create a force that equals or exceeds mg.
Let’s denote the additional compression required as d. When the spring is compressed by an additional distance d, the total compression becomes X + d. The spring force at this new compression is:
F_s = k * (X + d)
For the lower disc to just lift off the table, we need:
k * (X + d) = mg
Substituting the Spring Constant
Now, substituting the expression for k into this equation gives:
((2mg) / X) * (X + d) = mg
We can simplify this equation:
2mg * (X + d) = mg * X
Dividing both sides by mg (assuming m is not zero) leads to:
2(X + d) = X
Rearranging this gives:
2X + 2d = X
2d = -X
Thus, we find:
d = -X/2
Final Thoughts
This result indicates that to lift the lower disc off the table, the spring must be compressed further by half the initial compression distance X. However, since we cannot have a negative compression, this means that the spring must be extended by this amount after the initial compression is released. Therefore, the spring needs to be compressed an additional distance of X/2 to ensure that the lower disc is lifted off the table when the compressive force is removed.
