Understand the Force-Extension graph for a single spring.

6.

a)

Step 1: Understand the Force-Extension graph for a single spring.
For a spring obeying Hooke's law, the force ($F$) is directly proportional to the extension ($e$), so the graph of Force vs. Extension is a straight line passing through the origin, with a positive slope.

Step 2: Consider two identical springs connected end to end (in series).
When two identical springs are connected in series, the total extension for a given force is the sum of the extensions of each spring. If a force $F$ causes an extension $e$ in one spring, then the same force $F$ will cause an extension of $e$ in each of the two springs in series, resulting in a total extension of $2e$.
This means that for the same force, the total extension is doubled. Therefore, the effective spring constant of the series combination is half that of a single spring ($k_{series} = k/2$).

Step 3: Sketch the graph.
The new graph for two springs in series will also be a straight line passing through the origin. However, its slope will be half the slope of the graph for a single spring, indicating that it is less steep.

\begin{tikzpicture} \begin{axis}[ xlabel={Extension}, ylabel={Force (N)}, xmin=0, xmax=5, ymin=0, ymax=10, axis lines=middle, xtick=\empty, ytick=\empty, clip=false ] \addplot[thick, blue, domain=0:4] {2*x}; % Original spring (steeper) \node[above right, blue] at (axis cs: 3, 6) {Single spring}; \addplot[thick, red, domain=0:4] {x}; % Two springs in series (less steep) \node[above right, red] at (axis cs: 3, 3) {Two springs in series}; \end{axis} \end{tikzpicture}

(Description of sketch: Draw a graph with "Extension" on the x-axis and "Force (N)" on the y-axis. Draw two straight lines starting from the origin. The first line (representing a single spring) should have a steeper positive slope. The second line (representing two springs connected end to end) should have a less steep positive slope, approximately half the slope of the first line.)

b) Determine the extension of the system.

Step 1: Determine the extension of a single spring under a given force.
Given that each spring extends by $2 \text{ cm}$ when a force of $50 \text{ N}$ is suspended from it.
So, for one spring: $F_0 = 50 \text{ N} \implies e_0 = 2 \text{ cm}$.

Step 2: Analyze the series combination of springs C and D.
The $50 \text{ N}$ load is suspended from spring D, which is suspended from spring C. In a series connection, the force acting on each spring is the same as the total load.
Force on spring C ($F_C$) = $50 \text{ N}$.
Force on spring D ($F_D$) = $50 \text{ N}$.
Extension of spring C ($e_C$) = $2 \text{ cm}$ (since it's an identical spring under $50 \text{ N}$).
Extension of spring D ($e_D$) = $2 \text{ cm}$ (since it's an identical spring under $50 \text{ N}$).
The total extension of the series combination (C and D) is the sum of their individual extensions:

$$e_{CD} = e_C + e_D = 2 \text{ cm} + 2 \text{ cm} = 4 \text{ cm}$$

Step 3: Analyze the parallel combination of springs A and B.
The horizontal bar supports the combination of C and D, which carries the $50 \text{ N}$ load. Therefore, the total force acting downwards on the horizontal bar is $50 \text{ N}$.
This force is distributed equally between springs A and B because they are identical and connected in parallel.
Force on spring A ($F_A$) = $\frac{50 \text{ N}}{2} = 25 \text{ N}$.
Force on spring B ($F_B$) = $\frac{50 \text{ N}}{2} = 25 \text{ N}$.
Now, calculate the extension of spring A (or B) under a force of $25 \text{ N}$. Since extension is proportional to force:

$$$$