# Velocity released from a spring

## Homework Statement

Everything is on a frictionless horizontal surface. If a block is compressed against a spring for some distance x meters and then released, what is the velocity of the block the moment it loses contact with the spring?

## Homework Equations

W_s = -(1/2)kx^2 [if reference point is 0m]
W = (1/2)m(v_f)^2 - (1/2)m(v_i)^2

## The Attempt at a Solution

So if we take the compression of the spring to be our 0 point, then when it is released, the block travels x meters when it is fully released by the spring. The work done by the spring on the block is going to be negative. But since the initial velocity is 0, we get the equation: -(1/2)kx^2 = (1/2)m(v_f)^2. If you try to solve this for the velocity, you have to take the square root of a negative number. What am I doing wrong here?

Doc Al
Mentor
So if we take the compression of the spring to be our 0 point, then when it is released, the block travels x meters when it is fully released by the spring. The work done by the spring on the block is going to be negative.
Why do you think that the work done by the spring will be negative?

Which way does the spring push? Which way does the block move?

The displacement is in the same direction as the force of the spring pushing on it, so it should be positive, but then why does the textbook say that the work done by a spring is always -(1/2)kx^2 if the starting position is taken to be the 0 point?

Doc Al
Mentor
The displacement is in the same direction as the force of the spring pushing on it, so it should be positive, but then why does the textbook say that the work done by a spring is always -(1/2)kx^2 if the starting position is taken to be the 0 point?
The work done by the spring is negative when you compress it from 0 to x and positive when it decompresses from x to 0 (which is what is going on here).

I prefer to think in terms of the energy stored in the spring, which is simply 1/2kx^2.

It says that the work done is negative because the block or the system DOES work ON the spring AGAINST its restoring force.