Where does the half of Potential energy go? [duplicate]

We have the formula for elastic energy stored in a spring to be 1/2*Force*extension. When we have an object on a vertical spring, then the energy gained should be 1/2*mgh.(h is extension produced) The loss of Potential energy by the mass seems to be mgh. But shouldn't loss of potential energy by that object equal gain in elastic potential energy? Here, only half of original Potential energy is converted to elastic potential energy. Where did the rest of energy go?

marked as duplicate by Jon Custer, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 5 at 15:48

• – Farcher Apr 5 at 7:59

If I understand you correctly: you start with the spring in a position where there is no tension on the spring (the object is held in place by some other force), and then release the object . What you get is an oscillator: the object bounces up and down on the spring, oscillating around the point where the spring force balances gravity. Initially, as the object passes that point, the remaining energy will take the form of kinetic energy of the object. In real systems, that kinetic energy will eventually dissipate due to some sort of friction, until the object settles at the equilibrium point.

If the spring is extended by a distance $$x$$ from its equilibrium position then the energy stored in the spring is $$\frac 1 2 kx^2$$ where $$k$$ is the spring constant. So if you start with the spring unextended, then attach a mass $$m$$ to the spring and let the mass fall a distance $$h$$ such that $$kh=mg$$ then the energy stored in the spring is indeed $$\frac 1 2 kh^2 = \frac 1 2 mgh$$. The potential energy lost by the falling mass is $$mgh$$ and the remaining energy $$\frac 1 2 mgh$$ is the kinetic energy of the mass (assuming there is no friction or air resistance or other dissipative forces).

To check this, note that the acceleration of the mass after falling a distance $$x$$ is

$$a = g\left(1-\frac{x}{h}\right)$$

and $$\frac{d(v^2)}{dx} = 2v\frac{dv}{dx} = 2a$$

so $$v^2 = 2g\left(x-\frac{x^2}{2h}\right)$$

When $$x=h$$ we have

$$v^2 = 2g\left(h-\frac{h^2}{2h}\right)=gh \\ \Rightarrow \frac 1 2 mv^2 = \frac 1 2 mgh$$

• How do we get that a = g(1-x/h) part? – user68153 Apr 5 at 11:08
• @user68153 The net force on the mass is $F=mg-kx$, we know that $k=\frac{mg}{h}$ and $a=\frac{F}{m}$. – gandalf61 Apr 5 at 12:23