Deriving the formulae for energy

Question

We've recently started learning about work and energy and these were the definitions given to us which also lines up with a lot of stuff I've seen online, work is ''the dot product of force and displacement'' or ''it is the product of the magnitude of the force and the displacement along the direction of the force'', energy is ''the capacity/ability to do work''. Calculating work is pretty straight forward and it was said to us that ''energy is measured by the total work a body can do''. But in most places including my textbooks, lectures from my teachers I've seen the derivation of the equation for kinetic energy done like this:

Suppose a body at rest travels a distance $s$ in the direction of the force $F$ with acceleration $a$ and the final velocity of the body is $v$, the work done by the body will be $W=Fs=mas$ now from the equations of motion we have $v^2=u^2+2as$ and since $u$ i.e. the initial velocity is 0, we can solve for $as$ and it turns out to be $\frac{1}{2}v^2$ substituting this in the equation for work done we get $W=\frac{1}{2}mv^2$ and this is also the kinetic energy of the body, usually denoted by $E_k=\frac{1}{2}mv^2$

My issue with this approach is what we're doing here is we're calculating the amount of work done to reach the final position but according to the definition of energy and the way of measuring energy I've stated previously the kinetic energy of the body should be the amount of work the body can do before coming to rest, now if we take that approach,

Let's say the current velocity of the body is $v$ and it comes to rest after we apply a force $F$ which causes an acceleration, $a$(which is negative). Now the work that can be done by the body is again, $W=Fs=mas$ using the equations of motion again we have $0=v^2+2as$(since the final velocity is $0$ and the initial is $v$, solving for $as$ and subbing it in the equation for work we get $E_k=-\frac{1}{2}mv^2$, which is the result we wanted but this minus popped out unexpectedly, now my teacher suggested that I should've used the equation $v^2=u^2-2as$, to account for the retardation happening here, but I don't quite understand that, another hypothesis of mine was since the displacement is opposite to the direction of the force the work done would be $-|F||s|$ but that also leaves me with the same answer but I could've done some mistakes, nevertheless, I mainly have 2 questions left-

(i)Why am I not getting the right equation with this approach and

(ii)Why do so many places online and even my textbooks, teachers calculate the work done by a body to reach a position as the kinetic energy of a body in that position but the definition suggests that the kinetic energy of a body in a position is the amount of work a moving body can do until it comes to rest, could anyone please explain which of the approaches are actually correct and if the first approach I mentioned is indeed correct, why so? it seems to be different from the definition of energy, also I have also seen this in the derivation of the equation of gravitational potential energy for example but I decided to explain the scenario with kinetic energy since that was more common and made me more confused, sorry for the extremely long question, but it was a very specific question and I wanted to provide as much context as possible, thanks!

Answer 1

The correct answer is that we should not be using the Work Energy Theorem (WET) as a definition, and instead just postulate the energy, say kinetic energy. The reason for this is that if you try to study WET in the context of relativity, the derivation becomes to convoluted that nobody would want to present it as so.

I've seen the derivation of the equation for kinetic energy done like this:

[...]

My issue with this approach is what we're doing here is we're calculating the amount of work done to reach the final position but as according to the definition and way of measuring energy I've stated previously the kinetic energy of the body should be the amount of work the body can do before coming to rest

Your complaint is ok, but we should first accept that this is a valid computation if the aim is for this external work to be storing kinetic energy for later use.

now my teacher suggested that I should've used the equation $v^2=u^2−2as$, to account for the retardation happening here, but I don't quite understand that

Your teacher is correct. If you want to convert kinetic energy into other forms of energy, i.e. use the kinetic energy to do work, then the body is slowing down, and so you should be having, finally, $0=v^2-2as$ with $a>0$ and this formula that corresponds to the body slowing down to a stop. If you do that, then you will get the kinetic energy expression and work done expression all correct, without sign issues. Note that this is the opposite of the earlier textbook ``derivation", and you see that just as it is possible for external forces to store energy as kinetic energy, external forces can extract kinetic energy to do work.

Why am I not getting the right equation with this approach

If you take your teacher's suggestion, it will be correct.

Why do so many places online and even my textbooks, teachers calculate the work done by a body to reach a position as the kinetic energy of a body in that position but the definition suggests that the kinetic energy of a body in a position is the amount of work a moving body can do until it comes to rest, could anyone please explain which of the approaches are actually correct and if the first approach I mentioned is indeed correct, why so?

It is really because the concept of work, came many decades after the concept of kinetic energy and gravitational potential energy, and since they are all conserved and convertible into each other, either form is going to be fine and correct anyway.

That is, what you are trying to do is a bit too pedantic. Yes, of course if you want to stick to the definitions, then your teacher's suggestion to your approach will be the correct one. The textbook form is also correct, because in the end it gets the correct kinetic energy formula.

Answer 2

$$W_{net} = \Delta KE$$

In your first case, $$W_{net}=KE - 0$$ Here $W_{net}$ was +ve and so was $KE$.

In the second case, $$W_{net} = 0 - KE$$ $$KE = -(W_{net})$$ $$KE = \frac{1}{2} mv^{2}$$

In this case, work done by external agent,i.e, $W_{net}$ was negative.

Answer 3

I will start this answer with a general discussion, and then I will go to your question specifically.

There is the principle of relativity of inertial motion.

Example: let there be two trains, and they have a velocity relativity to each other. If you are a juggler, and you are performing inside a train carriage, can you tell from how the balls are moving on which train you are? You cannot; it feels the same. The only time you notice something is when the train undergoes a change of velocity.

A further implication of the principle of relativity of inertial motion is that while you can definitely tell when the train is changing velocity, whether you count that as an acceleration or a deceleration is arbitrary.

Whether you count a particular change of velocity as an acceleration or a deceleration follows from your choice of reference.

Of course, here on Earth the obvious choice is to use the entire Earth as reference of zero velocity. But the Earth has its daily rotation, and there is the motion of the Earth around the Sun, and there is the motion of our solar system around the center of mass of our Galaxy, etc.

My point is, distinction between acceleration and deceleration is not a measurable. What is measurable is how much change of velocity there is.

To your question specifically:

There is the amount of kinetic energy that must be transferred to an object in order to bring it up to a particular velocity $v$ relative to the reference coordinate system.

For a vehicle such as a car the obvious reference of velocity is the road. The speedometer of the car gives the velocity relative to the road.

There is the amount of kinetic energy that must be transferred to a car to bring it up to a particular velocity $v$.

Conversely, when you use the brakes to decelerate the car to a standstill the amount of kinetic energy that is transformed in that process is the same amount as it took to bring the car up to speed.

In both cases, acceleration and deceleration, the same amount of work is done, the difference is the direction of energy transfer.

During acceleration (relative to the road) the kinetic energy of the car (relative to the road) is increasing

During deceleration (relative to the road) the kinetic energy of the car (relative to the road) is decreasing

In the case of an electric car:
During acceleration (relative to the road): potential energy stored in the battery pack is transferred to the motors, the spinning motors drive the wheels, the wheels grip the road and the car is accelerated (relative to the road).

During deceleration there is the option of regenerative braking. The wheels grip the road, the wheels drive the motors, the motors are acting as generators, generating electric energy, and the state of charge of the battery pack increases.

(In actual cars regenerative braking does not recover all of the kinetic energy, since there are losses at every intermediate step, but the regenerative braking is certainly worthwhile.)

Answer 4

Work of the net force equals the change of kinetic energy, that is the relevant concept.

Considering rectilinear motion, when a body is accelerated from zero to a velocity $v$, the net force and displacement vectors are in the same direction, and the dot product is positive. Also, as you have shown, the change in kinetic energy is positive: $\frac{1}{2}mv^2$.

If the body is decelerated from a velocity $v$ to zero, the net force and displacement vectors have an angle of $180^{\circ}$, and the dot product is negative. Also the change in kinetic energy is negative: $-\frac{1}{2}mv^2$.

Answer 5

actually you have this 3 basic equations

case I

the force F is constant and positive , thus from Newton second law

$$m\,a=F\quad,\text{a is constant}$$ and the work W is $$W=F\,s=m\,a\,s\tag 1$$ $$v=u+a\,t\tag 2$$ $$s=u\,t+\frac 12 a\,t^2\tag 3$$

with $~u=0~$ you can solve the 3 equations for the unknows $~W~,s~,t~$ and obtain

$$W=\frac 12 \,m\,v^2$$

case II

the force F is constant and negative , thus from Newton second law

$$W=-F\,s=-m\,a\,s\tag 1$$ $$v=u+a\,t\tag 2$$ $$s=u\,t+\frac 12 a\,t^2\tag 3$$

again 3 equations with $~u=0~$ the result is

$$W=-\frac 12\,m\, v^2$$

but the minus sign is because the force is negative, the time t and the distance s are positive. equations (2) and (3) are equal in both cases.

this means that in case I and II you obtain $$v^2=u^2+2\,a\,s$$

Answer 6

This might be confusing, and if it is, then ignore it!

Suppose you are in a car and you go around a curve but there is no change in velocity as measured by your speedometer. To an observer on the ground, the tires are experiencing two forces working at right angles to each other: the weight (mass x acceleration due to gravity) of the car + a transverse force which is changing the directory of car. As the car continues through the curve, the tires continue to exert the transverse force. When the car is no longer changing direction, the tires no longer exert a transverse force. The tires still exert a force to counteract gravity. However, since you are inside the car, you experience a transverse force going in the opposite direction from the turn. This is because you are in an accelerated frame. Since the velocity of the car, as measured by your speedometer, has not changed, your kinetic energy has not changed. If your car was equipped with a gyro-compass (most cars are not), then you would know that the car had accelerated because you know that velocity is a vector.

That is one of the reasons why the formula for kinetic energy is ¹/₂ m v² is because squaring the velocity, which is another way of thinking about the dot product, gets rid of the vector quality.

Another way to think about this is imagine a canon, a canon ball, and a target. The canon does work on the canon ball, and the canon ball does work on what ever it hits. Note that the canon ball travels in a parabola. As it travels, some of its kinetic energy is converted to potential or gravitational energy. It slows. But once it is past the peak of the parabola, that potential energy converts back to kinetic energy. It speeds up.

I hope this makes things clearer.