Question

if the block is replaced by a block of mass 2m and allowed to hang at rest, how much will the spring be stretched compared to its equilibrium length? choose 1 answer: a 2gx b 2x c 4gx d 4x

Explanation:

Step1: Recall Hooke's Law and Equilibrium

For a spring with a mass \( m \) hanging at rest, the spring force \( F_s = kx \) (Hooke's Law) balances the weight \( F_g = mg \). So \( kx = mg \), which gives \( k=\frac{mg}{x} \).

Step2: Analyze for Mass \( 2m \)

When the mass is \( 2m \), let the new stretch be \( x' \). At rest, spring force balances weight: \( kx' = 2mg \). Substitute \( k=\frac{mg}{x} \) into this equation: \( \frac{mg}{x} \cdot x' = 2mg \).

Step3: Solve for \( x' \)

Simplify the equation: \( x' = \frac{2mg \cdot x}{mg}= 2x \).

Answer:

B. \( 2x \)