Work Done by Spring: Solving the Mystery

AI Thread Summary
The discussion centers on the work done by a variable force on a spring, specifically why the work done by an external force is not simply the product of the external force and the distance stretched. It is clarified that the force exerted by the spring varies as it is stretched, increasing with distance due to Hooke's law. As the spring stretches, it exerts a counteracting force that leads to instability until equilibrium is reached. The conversation highlights the dynamics of acceleration and deceleration of an object attached to the spring as it responds to the external force. Understanding these principles is essential for solving problems involving springs and variable forces.
T-Hau
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I have been scratching my head on this problem and hope that someone might solve my problem.
The question is about work done by a variable force on a spring. When a spring is stretched x distance from its initial position by an external force.Why the work done by the external force is not equal to F(external) * x?
 
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T-Hau said:
I have been scratching my head on this problem and hope that someone might solve my problem.
The question is about work done by a variable force on a spring. When a spring is stretched x distance from its initial position by an external force.Why the work done by the external force is not equal to F(external) * x?

Because the force is a variable and varies as a function of x.

Zz.
 
suppose, the natural position (where no force is acting on the spring) is at x=0 (stable). It's obvious that it will not stretch unless some external force is applied. Let us say, the action of any external force will lead to instability. Now, the spring exerts a force against the external force to attain stability. So, more the instability, more the force the spring is going to exert...
 
ZapperZ said:
Because the force is a variable and varies as a function of x.

Zz.

What u meant by function of x?
 
AudioFlux said:
suppose, the natural position (where no force is acting on the spring) is at x=0 (stable). It's obvious that it will not stretch unless some external force is applied. Let us say, the action of any external force will lead to instability. Now, the spring exerts a force against the external force to attain stability. So, more the instability, more the force the spring is going to exert...

Is it because as the spring is stretched more, it takes more force to stretch it?
 
T-Hau said:
Is it because as the spring is stretched more, it takes more force to stretch it?

as the spring is stretched more, more force is exerted by the spring on the body applying the external force.

if you apply a continuous and unchanging force F on an object attached spring whose other end is fixed to a wall, the object will first accelerate, till the acceleration becomes 0 (due to the backward force exerted by the spring). When the acceleration becomes 0, the object is not at rest but at a certain velocity. Now, it will start decelerating till the object come to rest.
 

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T-Hau said:
Is it because as the spring is stretched more, it takes more force to stretch it?

Are you familiar with Hooke's law? I'm a bit puzzled that you are asked to solve a problem using springs, but this appears to be a surprise to you.

Zz.
 
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