# Hooke's law & work by spring [duplicate]

Suppose a body is attached to a spring and a constant force $F$ is applied on the other side (Spring constant = $k$)at maximum extension the body is in equilibrium and so $$F=kx,$$ so we get maximum extension $x=F/k$, but if we take the work energy theorem approach then change in kinetic energy is zero as the body starts from rest and in the point of maximum extension comes to rest,so work done on it by all forces should be zero,work done by $$F=Fx$$ (as $F$ is constant), work done by spring = $$-\frac{k}{2}x^2,$$ since the work done on it must be zero, $$Fx-\frac{k}{2}x^2=0,$$ which gives us $$x=\frac{2F}{k}$$ which is different than the first answer,so which one is correct and which one is wrong?

• Why is $F$ constant? The spring force is a very good example of a variable conservative force. The force depends on the extension of the spring. Apr 18, 2017 at 15:35
• At maximum extension, the body is not in equilibrium. Apr 18, 2017 at 16:20
• Please change the title to be something more descriptive. Apr 18, 2017 at 17:16
• Possible duplicate of How much work can a force on a spring do? (Why are two methods wrong?) Apr 19, 2017 at 1:03

• constant force $F$ starts pushing body
• body reaches equilibrium point $x_e = \frac{F}{k}$ with zero velocity
This statements are contradictory. If constant $F$ pushes from the start, body is accelerating, since spring force can't compensate it. Body has positive acceleration until it reaches equilibrium point and only after it spring force becomes larger, so body becomes decelerating.
As you can see, body won't have zero velocity at equilibrium point. However, assumptions can be fixed: if $F$ is equal to spring force at any moment, body travels with constant speed. Thus no kinetic energy change is involved and you can find change in potential energy etc.