Question

if k = 20 n/m for the individual \springs\ in experiment #1, what is ( k_t ) for the system?
0.125 n/m
0.75 n/m
7.5 n/m
10.0 n/m
5.0 n/m

Explanation:

Step1: Recall Spring System Type (Assume Parallel or Series)

Assume Experiment #1 has springs in parallel? Wait, no—wait, maybe series? Wait, no, let's think. Wait, maybe the problem is about springs in parallel or series. Wait, but the options: let's assume the system is springs in parallel? No, wait, maybe series? Wait, no, let's check the options. Wait, maybe the system has 4 springs? Wait, no, let's think again. Wait, maybe the problem is about springs in parallel? Wait, no, let's recall: for springs in parallel, \( k_T = k_1 + k_2 + ... \); for series, \( \frac{1}{k_T} = \frac{1}{k_1} + \frac{1}{k_2} + ... \). Wait, but the options: 5.0 N/m? Wait, maybe the system has 4 springs? Wait, no, let's see: if \( k = 20 \) N/m, and if the system is 4 springs in series? Wait, no, \( \frac{1}{k_T} = \frac{4}{20} = \frac{1}{5} \), so \( k_T = 5 \) N/m. Oh, that matches one of the options (5.0 N/m). So likely, Experiment #1 has 4 springs in series.

Step2: Calculate \( k_T \) for Series Springs

For \( n \) identical springs in series, \( \frac{1}{k_T} = \frac{n}{k} \), so \( k_T = \frac{k}{n} \). Wait, if \( n = 4 \), then \( k_T = \frac{20}{4} = 5 \) N/m. That's one of the options (5.0 N/m).

Answer:

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