# Work done of a spring.

1. Jul 25, 2011

### T-Hau

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?

2. Jul 25, 2011

### ZapperZ

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

Zz.

3. Jul 25, 2011

### AudioFlux

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...

4. Jul 25, 2011

### T-Hau

What u meant by function of x?

5. Jul 25, 2011

### T-Hau

Is it because as the spring is stretched more, it takes more force to stretch it?

6. Jul 25, 2011

### AudioFlux

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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7. Jul 25, 2011

### rcgldr

8. Jul 25, 2011

### ZapperZ

Staff Emeritus
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.