# Work done by spring force

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?

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

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