I have a problem with this question I'm attempting. It says an electron has kinetic energy $K_1$ and work is done until its moving in the opposite direction with a quarter of its original velocity. I've solved it to find the work done was $-15K_1/16$ but it then asks me if the answer depends on the electrons direction. So I thought, if the electron was moving in its original direction instead of the opposite direction, then its velocity would be reduced by 75% but in the case above its velocity has been changed by 125% (from positive $V_1$ to negative $V_2/4$ so more work done should be done to move it in the opposite direction but the textbook says no. Why is this?

Edit: This is how I got the value of work done.

$W = K_2 - K_1$

$W = m/2 ((V_1/4)^2-V_1^2)$

$W = -15mV_1^2/32 = -15K_1/16$

  • $\begingroup$ Since this is a homework problem, you should show us a little more work. Like, how did you get $-15K_1/16$? Then, ask yourself if any of the decisions you made when solving the problem depended on the direction of the electron. I'll also point out that I feel the problem is slightly unclear - is it asking about the initial direction, the final direction, or the fact that the electron changed direction at all? We can answer all of them, but we should see how you got the answer you got first. $\endgroup$ – levitopher Nov 8 '19 at 22:13
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    $\begingroup$ Added to edit. I can see $V_1/4$ would be the same regardless of the direction since kinetic energy uses magnitude of the velocity. So work done is same in reducing the velocity by 75% as it is for 125%? $\endgroup$ – user552217 Nov 8 '19 at 22:45
  • $\begingroup$ Yeah, see Pablo's answer below. Kinetic energy only depends on the speed, so the direction doesn't matter. You gotta take some energy out of the system to slow it down, and if you slow it to zero you might have to add some back in, but it's always the same final amount of energy if the speed is fixed. $\endgroup$ – levitopher Nov 9 '19 at 3:10

According to the Work-Energy Theorem, the net work done to some object is equal to the difference in its kinetic energy, this is, \begin{align} W_{net}=K_{final}-K_{initial}. \end{align}

In order to understand what you mean by "independent of velocity direction", suppose we have an object at rest, and you can move it either to the right or to the left. Firstly, consider that we move it to the right by some force so it ends up with a velocity $v$, then the net work done is \begin{align} W_{net}&=\frac{1}{2}mv^2 - 0. \\ &=\frac{1}{2}mv^2 . \end{align} On the other hand, if we move it to the left so it ends up with a velocity $v_1$ but in the opposite direction, then the total work is the same as before, this is because in both cases its initial state is rest, and its final state have the same kinetic energy. This means that net work $\textbf{doesn't care at all}$ in which direction the object is moving, but instead in its initial and final $\textbf{velocity magnitude}$ .

Going back to your problem, what happens is that we apply some work $W_1$ to decelerate the body down to 25% its initial velocity, so the object's kinetic energy will be equal to $K_1 /16$. This way, we get \begin{align} W_1 &= K_1 / 16 - K_1 \\ &= -\frac{15}{16}K_1. \end{align} Now, if we want to reverse the object's velocity direction as you point out, then we need to exert some work $W_2$ to slow it down to rest, and afterwards we exert an additional work $W_3$ to accelerate it up to the a final kinetic energy equal to $K_1 / 16$, so it moves in the opposite direction. We can note that $W_2$ takes the body initially at $K_1 / 16$ to a final state of rest, and $W_3$ takes the body initially at rest to a final state with $K_1 / 16$ kinetic energy. Hence, $W_2$ and $W_3$ have the same magnitude but opposite sign, this is, \begin{align} W_2 &= 0 - K_1 / 16 \\ &= - K_1 / 16, \\ W_3 &= K_1 / 16 - 0 \\ &= K_1 / 16. \end{align} This way, the total work exerted upon the object in order to reverse its direction after the work $W_1$ has been done is equal to $W_2 + W_3 = 0$. This show up the fact that the direction of movement in the final state of the object is irrelevant in terms of the net work done to it.


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