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.

two figures: figure 1 shows a spring with length ( l ) holding a mass ( m ); figure 2 shows the same spring with length ( \frac{3l}{4} ) without the mass.

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{13l}{4} )
d. ( 4l )

Explanation:

Step1: Find the spring constant \( k \)

The unstretched length of the spring is \( \frac{3L}{4} \), and when a mass \( m \) is hung, the stretched length is \( L \). So the extension \( x_1 \) for mass \( m \) is \( L - \frac{3L}{4}=\frac{L}{4} \).
Using Hooke's law \( F = kx \), and at equilibrium, the force \( F \) is the weight \( mg \). So \( mg=k\times\frac{L}{4} \), which gives \( k = \frac{4mg}{L} \).

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

For a mass \( 4m \), the weight is \( 4mg \). Let the extension be \( x_2 \). Using Hooke's law \( 4mg=k x_2 \). Substitute \( k=\frac{4mg}{L} \) into this equation:
\( 4mg=\frac{4mg}{L}\times x_2 \)
Divide both sides by \( 4mg \) (assuming \( mg
eq0 \)): \( 1=\frac{x_2}{L} \), so \( x_2 = L \).

Step3: Find the new length of the spring

The unstretched length is \( \frac{3L}{4} \), and the extension for mass \( 4m \) is \( L \). So the new length \( l_{new}=\frac{3L}{4}+L=\frac{3L + 4L}{4}=\frac{7L}{4} \)? Wait, no, wait. Wait, earlier calculation of \( x_1 \): Wait, when mass \( m \) is hung, the length is \( L \), unstretched is \( \frac{3L}{4} \), so extension \( x_1 = L-\frac{3L}{4}=\frac{L}{4} \). Then \( mg = kx_1\Rightarrow k=\frac{mg}{x_1}=\frac{mg}{\frac{L}{4}}=\frac{4mg}{L} \). Then for mass \( 4m \), \( F = 4mg=kx_2\Rightarrow x_2=\frac{4mg}{k}=\frac{4mg}{\frac{4mg}{L}} = L \). Then the new length is unstretched length plus extension: \( \frac{3L}{4}+L=\frac{3L + 4L}{4}=\frac{7L}{4} \)? But wait, the options have \( \frac{7L}{4} \) as option a? Wait no, wait the options: a is \( \frac{7L}{4} \)? Wait the original problem's options: let me check again. Wait the user's options: a is \( \frac{7L}{4} \)? Wait the user wrote: a. \( \frac{7L}{4} \)? Wait the user's options: "a. \( \frac{7L}{4} \) b. \( 3L \) c. \( \frac{13L}{4} \) d. \( 4L \)". Wait my mistake in step 3: Wait, no, wait the unstretched length is \( \frac{3L}{4} \), extension for \( 4m \) is \( L \), so new length is \( \frac{3L}{4}+L=\frac{3L + 4L}{4}=\frac{7L}{4} \)? But wait, let's re - examine the first step. Wait, maybe I messed up the unstretched length. Wait the problem says: "the spring returns to its unstretched length of \( \frac{3L}{4} \)" (wait the user's original text: "unstretched length of \( \frac{3L}{4} \)". So when mass \( m \) is hung, length is \( L \), unstretched is \( \frac{3L}{4} \), so extension \( x_1 = L-\frac{3L}{4}=\frac{L}{4} \). Then \( mg = kx_1\Rightarrow k = \frac{mg}{x_1}=\frac{mg}{\frac{L}{4}}=\frac{4mg}{L} \). Then for mass \( 4m \), \( F = 4mg=kx_2\Rightarrow x_2=\frac{4mg}{k}=\frac{4mg}{\frac{4mg}{L}}=L \). Then new length is \( \frac{3L}{4}+x_2=\frac{3L}{4}+L=\frac{7L}{4} \), which is option a. Wait but let me check again. Wait, maybe I made a mistake in the unstretched length. Wait the problem says: "the spring returns to its unstretched length of \( \frac{3L}{4} \)". So when the block of mass \( m \) is hanging, the length is \( L \), so the extension is \( L-\frac{3L}{4}=\frac{L}{4} \). Then Hooke's law: \( mg = k\times\frac{L}{4} \), so \( k = \frac{4mg}{L} \). Then for mass \( 4m \), the force is \( 4mg \), so \( 4mg=k\times x \), so \( x=\frac{4mg}{k}=\frac{4mg}{\frac{4mg}{L}} = L \). Then the new length is unstretched length (\( \frac{3L}{4} \)) plus extension (\( L \)): \( \frac{3L}{4}+L=\frac{3L + 4L}{4}=\frac{7L}{4} \). So the answer should be option a. \( \frac{7L}{4} \).

Answer:

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