Physics – Why is Work Done in a Spring Positive?

forcespotentialpotential energyspringwork

We know that a stretched spring obeys Hooke's law, such that $F=-kx$.

We can find the potential energy of stretching/compressing this spring by $x$, given by :

$$U_x-U_0=-\int_0^x F.dx = \frac{1}{2}kx^2 $$

Setting $U_0=0$ as reference, we have $U_x=\frac{1}{2}kx^2$

However, this is also sometimes described as the work done by the spring.

Shouldn't the work done $W$ be given by $\int F.dr$, such that $W=-\Delta U = -\frac{1}{2}kx^2$ in this case ?

Isn't the work done by the spring negative ?

Also, in this case the potential energy comes to be negative.. In general, can we set any point as reference and set it to be $0$ and perform the integral between any two limits, to get either a positive or a negative $U$ ?

For example, in forces of the nature $r^{-n} ,(n>1)$ we usually take the reference at $r=\infty$ and integrate from $\infty$ to some point $r$. In case of forces of the nature $r^n$, we usually take $0$ as the reference and integrate from $0$ to some $r$. In general, we are free to choose any reference and any limit, even though some are much more convenient, right ? In theory, we can choose any point, right ?

As long as we have :

$$U_a-U_b=-\int_b^a F.dx$$ we can choose any $a$ and $b$, and set either of $U_a$ or $U_b$ to be the reference and equal to $0$, right ?

Best Answer

Setting $x=0$ as the reference point means you are looking at the work done by the spring from $x=0$ to the end position $x$. Since $W=-\Delta U=-\frac12kx^2$, this will always be negative, which makes sense since the spring force always points towards $x=0$, and thus will point opposite the displacement.

In general $$W_{a\to b}=-(U(x_b)-U(x_a))=\frac12k(x_a^2-x_b^2)$$ and this is positive whenever $x_a^2>x_b^2$

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