---
title: "A block of mass \\( m \\) is attached to the bottom of a light vertical spring of spring constant \\( k \\) that is suspended from a ceiling. The block is gently lowered until it hangs at rest at its equilibrium position. The spring stretches a vertical distance \\( y_0 \\) from its unstretched length to reach this equilibrium position. The downward direction is defined as positive, and \\( y = 0 \\) is defined as the position of the block when the spring is at its unstretched length."
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url: "https://nerd-notes.com/ubq/111613/"
date_modified: "2026-04-20T09:11:59+00:00"
---

# A block of mass \( m \) is attached to the bottom of a light vertical spring of spring constant \( k \) that is suspended from a ceiling. The block is gently lowered until it hangs at rest at its equilibrium position. The spring stretches a vertical distance \( y_0 \) from its unstretched length to reach this equilibrium position. The downward direction is defined as positive, and \( y = 0 \) is defined as the position of the block when the spring is at its unstretched length.

A block of mass \( m \) is attached to the bottom of a light vertical spring of spring constant \( k \) that is suspended from a ceiling. The block is gently lowered until it hangs at rest at its equilibrium position. The spring stretches a vertical distance \( y_0 \) from its unstretched length to reach this equilibrium position. The downward direction is defined as positive, and \( y = 0 \) is defined as the position of the block when the spring is at its unstretched length.

![Three side-by-side vertical configurations showing a spring attached to a ceiling. Left: An unstretched spring with its bottom end aligned with a horizontal dashed line labeled 'Unstretched length, y = 0'. Middle: The same spring stretched with a block of mass 'm' attached. The block's center aligns with a second horizontal dashed line labeled 'Equilibrium position, y = y_0'. Right: The spring is stretched further downward with the same block 'm'. The block's center aligns with a third horizontal dashed line labeled 'Release position, y = y_0 + D'. A vertical axis arrow points downward with a '+y' label.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1776122421-IADbRl.jpg)

**Part a)** **Derive** an expression for the equilibrium stretch distance \( y_0 \). Express your answer in terms of \( m \), \( k \), and physical constants, as appropriate. *(2 points)*

**Part b)** The block is now pulled down an additional distance \( D \) from the equilibrium position and released from rest at time \( t = 0 \). Starting from the conservation of energy principle, **derive** an expression for the maximum speed \( v_{\text{max}} \) of the block during its oscillation. Express your answer in terms of \( D \), \( k \), \( m \), and physical constants, as appropriate. Show all steps of your derivation, including the terms for both gravitational potential energy and elastic potential energy. *(4 points)*

**Part c)** The original block is removed and replaced by a new block of mass \( 2m \). The new block is gently lowered to its own equilibrium position, then pulled down by the *same* additional distance \( D \) and released from rest. *(4 points)*

**Part d)** On the axes provided, **sketch** a graph of the total elastic potential energy of the spring \( U_s \) as a function of the block's vertical position \( y \). Clearly label the positions \( y_0 \) and \( y_0 + D \) on the horizontal axis. *(2 points)*


*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/111613/*