Question

if $k = 20$ n/m for the individual \springs\ in experiment #1, what is $k_t$ for the system?

• $7.5$ n/m
• $0.125$ n/m
• $0.75$ n/m
• $10.0$ n/m
• $5.0$ n/m

Explanation:

Step1: Recall Spring System Type

Assume Experiment #1 has springs in parallel or series. For series, $\frac{1}{k_T}=\sum\frac{1}{k_i}$; for parallel, $k_T=\sum k_i$. But common experiment #1 (like 4 springs? Wait, maybe 4 springs in series? Wait, no—wait, maybe it's a system with 4 springs? Wait, no, let's think. Wait, if individual k=20 N/m, and let's suppose the system is 4 springs in series? Wait, no, the options are 5,10, etc. Wait, maybe it's 2 springs in parallel? No, wait, maybe 4 springs? Wait, no, let's check the options. Wait, maybe the system is 4 springs in series? Wait, no, 1/20 +1/20 +1/20 +1/20 = 4/20 = 1/5, so k_T=5? Wait, that matches one option. Wait, maybe Experiment #1 has 4 springs in series? Let's verify.

Step2: Calculate for Series (4 springs)

If there are n springs in series, $k_T = \frac{k}{n}$ (if all k same). Wait, no: $\frac{1}{k_T} = \frac{n}{k}$, so $k_T = \frac{k}{n}$. If n=4, k=20, then $k_T = 20/4 = 5$ N/m. That matches the option 5.0 N/m.

Answer:

5.0 N/m (the option with 5.0 N/m)