0
$\begingroup$

Suppose an ideal spring is attached to a wall at one of its end. Let an external force act on the spring at another end to stretch the spring to distance $x$. If spring constant is $k$ then work done by external force will be $(1/2)kx^2$.

The above expression for work done by external force is arrived using integration. Let us roll back to definition of work done. Work is said to be done on an object if 1) force is applied on the object 2) displacement of object happens

My question is that in the above spring the displacement of spring (due to external force) as a whole has not taken place. Instead it is just "stretched". How can we say displacement of object(spring) happened and work is done.

The same criteria also applies to all elastic bodies. For example ● work done in compressing an ideal gas($PdV$ work) ● work done in extending a wire having young modulus Y( $(AYl^2)/(2L) $ )

I do not know what is the correct way to define displacement! Please help me in understanding it.

$\endgroup$
  • $\begingroup$ think of Newton's third law in that context. $\endgroup$ – image Mar 28 '15 at 12:10
  • $\begingroup$ Well, streching a spring does makes something move, right? You can not compress or stretch anything without changing it's lenght / volume. This change is a displacement, at least of some particles in the object. $\endgroup$ – Steeven Mar 28 '15 at 16:17
1
$\begingroup$

If the spring has been stretched then something must have stretched it. It's the work done by that something that you're calculating. That work has been done on the spring, so the potential energy of the spring has increased by an amount equal to the work done on it.

To give a concrete example, suppose you're holding the end of the spring. Then your hand exerts a force $F$ and your hand moves a distance $x$. The work done by your hand is given by:

$$ W = \int_0^L F(x)dx $$

where $L$ is the displacement of your hand. For an ideal spring $F = kx$ giving us the expression for the work done as a function of the displacement of your hand:

$$ W = {1\over2}kL^2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.