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Body of mass $1\rm\, kg$ is moving towards left side at velocity of $-2\rm\,m/s$ and is slowing down due to the constant external force until it stops. After that, it's speeding up towards the right side due to the same force until it reaches velocity of $2\rm\,m/s$. How much work has the force done on the body?

Correct answer is 0 Joules and we get that if we use formula $E_k=\frac 1 2 mv_f^2 - \frac 1 2 mv_i^2$. But that makes no sense to me. Basically, an object was stoped and accelerated without work. How is that possible?

It would have more sense for me if it was 8 Joules because $E_k= \frac 1 2 mv^2 = \frac{1\rm\, kg\cdot 16\rm\, m^2/s^2}{2} = 8\rm\, J$.

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    $\begingroup$ Net work is 0, this doesn't mean 0 work has been done through a partition of the journey $\endgroup$ Commented Oct 8, 2022 at 11:44

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Suppose we apply a uniform force $F$, which is of course positive. Now let's say that the total work $W$ is the sum of the work when the body moves to the left side $W_-$ and when the body moves to the right side $W_+$: $$W=W_-+W_+$$ By definition of work, $$W=W_-+W_+= F\cdot \Delta x_-+ F\cdot \Delta x_+= F\cdot (\Delta x_-+\Delta x_+)$$ As can be seen from the graph $v(t)$, if the body starts at $x=x_{0}$, it ends at $x=x_{0}$ as well. $$W=F\cdot (\Delta x_-+\Delta x_+)=F\cdot ((x_{min}-x_{0})+(x_{0}-x_{min}))=0$$ where we used the definition of the displacement $\Delta x=x_{final}-x_{initial}$. Therefore, the work is 0, indeed.


The fundamental idea is that at first the force points to the other direction of the displacement (negative work), and then points to the same direction as the displacement (positive work). At some point in time, the total energy of the system is smaller than the initial, but in the end it is the same. The work was done, but the net work is zero.

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