Question

a group of students is experimenting with two springs. spring 1 and spring 2. spring 1 stretches by a distance of 0.400 m when a block weighing 8.0 n is hung vertically from it. when the same block is hung vertically from spring 2, spring 2 stretches by a distance of 0.100 m. the two springs are then connected in series, and a smaller 4.0 n block is hung vertically from this combination of springs. what is the distance that this combination will stretch from its unstretched length? a 0.040 m b 0.080 m c 0.125 m d 0.250 m

Explanation:

Step1: Calculate spring constants

Using Hooke's law \(F = kx\), for Spring 1: \(k_1=\frac{F_1}{x_1}\), where \(F_1 = 8.0\space N\) and \(x_1=0.400\space m\). So \(k_1=\frac{8.0}{0.400}=20\space N/m\).
For Spring 2: \(k_2=\frac{F_2}{x_2}\), where \(F_2 = 8.0\space N\) and \(x_2 = 0.100\space m\). So \(k_2=\frac{8.0}{0.100}=80\space N/m\).

Step2: Calculate equivalent spring constant for series combination

The formula for equivalent spring constant (\(k_{eq}\)) of two springs in series is \(\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\).
Substituting \(k_1 = 20\space N/m\) and \(k_2=80\space N/m\), we get \(\frac{1}{k_{eq}}=\frac{1}{20}+\frac{1}{80}=\frac{4 + 1}{80}=\frac{5}{80}=\frac{1}{16}\). So \(k_{eq}=16\space N/m\).

Step3: Calculate total stretch for \(4.0\space N\) block

Using \(F = k_{eq}x_{total}\), where \(F = 4.0\space N\) and \(k_{eq}=16\space N/m\). Then \(x_{total}=\frac{F}{k_{eq}}=\frac{4.0}{16}=0.250\space m\).

Answer:

D. \(0.250\space m\)