- #1

blalien

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**[SOLVED] Mass on a spring**

## Homework Statement

This problem is from Gregory’s Classical Mechanics

A light spring of natural length a is placed on a horizontal floor in the upright position. When a block of mass M is resting in equilibrium on top of the spring, the compression of the spring is a/15. The block is now lifted to a height 3a/2 above the floor and released from rest. Find the compression of the spring when the block first comes to rest.

## Homework Equations

Mgh = the potential energy of the mass at a height h above the floor

1/2kx^2 = the potential energy of the spring, since we are assuming that the spring obeys Hooke's Law

## The Attempt at a Solution

It doesn’t help that the book never defines “natural length” or “compression.” But anyway, this is my logic:

Assume that, on the horizontal floor, V = 0. So, you have two systems and two equations:

Mass starts at rest on spring -> Mass compresses spring by a/15

Mga = Mg(14a/15)+1/2k(a/15)^2

Mass starts at 3a/2 above floor -> Mass compresses spring by x

Mg(3a/2) = Mg(a-x)+1/2kx^2

Then you solve for k and x. Unfortunately, this yields the wrong answer, which is apparently x = a/3. I don’t want to be told exactly how to do the problem. But, could you please just point to the faulty part in my logic? I would really appreciate the help.

Thanks!