Question

1. a block of mass m is hanging in equilibrium from a spring. the block stretches the spring to length l, as shown in figure 1. the block is removed, and the spring returns to its unstretched length of \\(\frac{3l}{4}\\) as shown in figure 2.

figure 1: spring with mass m, length l; figure 2: unstretched spring, length \\(\frac{3l}{4}\\)

if a larger block with mass 4m is hung from the spring and slowly lowered until it is in equilibrium, what will the new length of the spring be?

a. \\(\frac{7l}{4}\\)
b. 3l
c. \\(\frac{15l}{4}\\)
d. 4l

Explanation:

Step1: Find the spring constant \( k \)

When the mass \( m \) is hanging, the extension of the spring is \( x_1 = L - \frac{3L}{4}=\frac{L}{4} \). At equilibrium, the spring force \( F = kx_1 \) balances the weight \( mg \), so \( mg = k\frac{L}{4} \), which gives \( k=\frac{4mg}{L} \).

Step2: Find the extension for mass \( 4m \)

For mass \( 4m \), let the extension be \( x_2 \). At equilibrium, \( 4mg = kx_2 \). Substitute \( k=\frac{4mg}{L} \) into this equation: \( 4mg=\frac{4mg}{L}x_2 \). Solving for \( x_2 \), we get \( x_2 = L \).

Step3: Find the new length of the spring

The unstretched length is \( \frac{3L}{4} \), so the new length \( L_{new}=\frac{3L}{4}+x_2 \). Substitute \( x_2 = L \) into it: \( L_{new}=\frac{3L}{4}+L=\frac{3L + 4L}{4}=\frac{7L}{4} \)? Wait, no, wait, there is a mistake. Wait, when mass \( m \), extension is \( L-\frac{3L}{4}=\frac{L}{4} \), so \( mg = k\frac{L}{4} \), so \( k = \frac{4mg}{L} \). For mass \( 4m \), \( 4mg=kx_2 \), so \( x_2=\frac{4mg}{k}=\frac{4mg}{\frac{4mg}{L}} = L \). Then new length is unstretched length \( \frac{3L}{4} \) plus extension \( x_2 = L \), so \( \frac{3L}{4}+L=\frac{3L + 4L}{4}=\frac{7L}{4} \)? But wait, option a is \( \frac{7L}{4} \)? Wait, no, let's check again. Wait, maybe I made a mistake in unstretched length. Wait, Figure 2 shows unstretched length is \( \frac{3L}{4} \), Figure 1: length is \( L \), so extension is \( L - \frac{3L}{4}=\frac{L}{4} \). So for mass \( m \), \( mg = k\frac{L}{4} \), so \( k = \frac{4mg}{L} \). For mass \( 4m \), force is \( 4mg \), so extension \( x = \frac{4mg}{k}=\frac{4mg}{\frac{4mg}{L}} = L \). Then new length is \( \frac{3L}{4}+L=\frac{7L}{4} \), which is option a? Wait, option a is \( \frac{7L}{4} \)? Wait, the options: a. \( \frac{7L}{4} \), b. \( 3L \), c. \( \frac{15L}{4} \), d. \( 4L \). So yes, the calculation gives \( \frac{7L}{4} \), which is option a.

Answer:

a. \( \frac{7L}{4} \)