Maximum velocity by compressing spring

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Homework Help Overview

The discussion revolves around a physics problem related to the maximum velocity of an object launched by a compressed spring. The original poster, who has a programming background, is exploring the relationship between kinetic energy and potential energy in the context of spring mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate kinetic energy and potential energy in their equations, questioning how to express the maximum velocity of an object when released from a compressed spring. Some participants clarify the distinction between launching an object and slowing it down against a spring, while others provide insights into energy conservation principles.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the energy relationships involved. The original poster has adjusted their approach based on feedback, and there is ongoing exploration of how to derive the maximum velocity from the energy equations discussed.

Contextual Notes

There is some confusion regarding the setup of the problem, particularly whether the object is being launched or slowed down by the spring. This ambiguity may affect the interpretation of energy conservation in the scenario.

badsniper019
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Homework Statement


I'm new to physics. I'm a programmer, so I'm having trouble finding this solution. when spring is compressed against wall until the body attached to it touches the wall, what's the maximum velocity obtained?


The Attempt at a Solution


I thought it happens when the kinetic energy is high. Of the original kinetic energy of the body, some goes into potential energy in the spring and the remainder is left as kinetic energy of the now slowed object. So I did the remainder of energy is equal to force(-kx) over time, which means
\frac{1}{2}(mv^2-kX^2)=-kXt

x is the distance obtained, t is time
 
Last edited:
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Welcome to PF;
This is homework?

You seem to have written you problem two ways - the title, and initial comments, appear to want to set an object moving by launching it from a compressed spring... but your working suggests you are slowing an object by running it against a spring.

In the first case, the energy stored in the compressed spring turns into kinetic energy in the object (under ideal conditions - in real life, some of it goes into kinetic energy in the spring - because a real spring has mass.)

In the second - it is the reverse, only you may have some kinetic energy left over (in which case the object slams into the wall as well.)

If the speed is such that the object comes to rest after compressing the spring a distance x, then all the initial kinetic energy gets stored in the spring (well - ideally - IRL there will be some losses) ... nothing left over.
 
sorry about that. I'm working on the first case. So my equation should be 1/2(mv^2)=-kxt?
 
sorry about that. I'm working on the first case. So my equation should be 1/2(mv^2)=-kxt?
 
Energy stored in a spring compressed by distance x is: ##P=\frac{1}{2}kx^2## ... you can look these things up :)
 
Ok, got it. 1/2(mv^2)=1/2(kx^2). which gives v=\sqrt{\frac{k}{m}}x. But how to find the maximum velocity. Derivative over velocity?
 
You have just worked out the speed if all the energy stored in the spring goes into the motion of the object. That's the biggest it can get.
 
So this is the formula for maximum velocity?
 
If you can think of a way to get more kinetic energy than the total PE stored in the spring, let me know.
 

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