0
$\begingroup$
  1. I wanted to clarify if kinetic energy is always positive. Since $KE = 1/2 mv^2$, and $m$ and the square of $v$ is positive. I assume as such.

  2. Given that I have a scenario where an object which was travelling at a positive velocity in a certain direction (we take this direction as positive), reverses the direction and travels at a negative velocity similar in magnitude. The $KE$ final and initial would be equal due to $KE$ always being positive. Now, the change in $KE$ would be $0$, since both values are equal positive values. However, that is not the case as $W = Fd\cos(\theta)$. And clearly, there must be a force enacted to change the velocity of the object. Assume there is a distance present over which the force acts. Now, the work should be negative as $\cos(\theta)$ is negative.

Given this, I don't understand what I am supposed to do when writing $KE$ or $W$. If $KE$ is negative relative to something else, would I write $KE = - 1/2 mv^2$?

  1. $W=Fd\cos(\theta)$ or $W=Fs$. In the $\cos\theta$ version, are $F$ and $d$ magnitude/s, vector/s or scalar/s? In the $W=Fs$ version, I have the same question.

  2. I read somewhere that scalars are simply values. In that case, are scalars magnitudes with signs, rather than just magnitudes? Or is a magnitude being an absolute value a misconception?

  3. Say that I have a velocity time graph, and the velocity changes at a certain point. These points of change can not be differentiated can they? Would the end point's slope be undefined as well?

$\endgroup$
  • $\begingroup$ Too many different questions at once. Some (perhaps all) are already answered by other questions on this site. Please check before posting a new question. $\endgroup$ – sammy gerbil Mar 28 at 2:04
1
$\begingroup$

W = fdcos(theta). And clearly, there must be a force enacted to change the velocity of the object. Assume there is a distance present over which the force acts. Now, the work should be negative as cos(theta) is negative.

Here is where you go wrong. $\cos(\theta)$ is negative in the beginning as the object slows from the initial positive velocity to 0. At that point the object reversed direction. From then on $\cos(\theta)$ is positive and the work is positive during the time while it goes from 0 to the final negative velocity. Over the entire path the work is zero, the positive work in the last part being equal to the negative work in the first part.

The change in KE is zero as is the work.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.