Conundrums on energy and graph [closed]

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?

• 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. – sammy gerbil Mar 28 at 2:04

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.