Velocity released from a spring

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    Spring Velocity
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Homework Help Overview

The problem involves a block released from a compressed spring on a frictionless horizontal surface, with the goal of determining the block's velocity at the moment it loses contact with the spring. The discussion centers around the concepts of work done by the spring and energy transformations in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the work done by the spring and the direction of forces involved. Questions arise regarding the sign of the work done by the spring and its implications for the energy equations being used.

Discussion Status

Participants are actively questioning the assumptions regarding the work done by the spring, particularly its sign and how it relates to the motion of the block. Some guidance is provided regarding the interpretation of work in terms of energy stored in the spring.

Contextual Notes

There is a focus on the definitions and conventions used in the problem, particularly concerning the reference point for work and the implications of the spring's restoring force. The discussion reflects uncertainty about the application of energy equations in this specific scenario.

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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?
 
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PhizKid said:
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?
 
PhizKid said:
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.
 

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