1
$\begingroup$

Consider a spring connected to blocks on it's ends lying on smooth horizontal table. Now let the right end block be displaced $x_1$ and left end be displaced $x_2$ from the mean position such that the work done by spring is $ -\frac{1}{2}k(x_1+x_2)^2$ (=$-\Delta PE$) But my question is if we consider the the free body of a single block then $F=-kx$ acts on the block and the work done by that force is $-\frac{1}{2}k(x_1+x_2)^2$ (I understand why work done on say right block is not $(-\frac{1}{2}kx_1)^2$ as if we consider right end block at any instant the ‘$x$’ term in the force is total extension of the spring at that instant) and as there are two blocks so total work is $=2\cdot(\frac{1}{2}k(x_1+x_2)^2)$ Where am I going wrong?

$\endgroup$
  • $\begingroup$ How could a spring connect to one block in opposite sides. Or is the spring is curved? $\endgroup$ – Sandesh Goli Oct 13 '19 at 10:26
1
$\begingroup$

In the scenario you are considering it is no longer true that $F_1=-k x_1$ because the other side may move and change the force irrespective of $x_1$. So $F_1=-k (x_1+x_2)$.

When calculating the work done by $F_1$ you have to include both $x_1$ and $x_2$ in the calculation of the force, but only $x_1$ in the calculation of the distance. Similarly for $F_2$.

So when you calculate the work done by the individual forces you get a complicated function that depends on their joint motion. In the end, however, you will always find that they add up to $-\Delta PE$, but you can make the work done by either individual force take any value you want by appropriately moving the other end.

$\endgroup$
  • $\begingroup$ So do you mean work done on say right block when it is displaced by x is $\int_0^xk(x_1+x_2)dx_1=-1/2kx_1^2 -kxx_2$ $\endgroup$ – Srinivas K Oct 14 '19 at 0:41

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.