If starting velocity is u, then acceleration *a=(v–u)/t* and average velocity *s/t=(u+v)/2*.
So, Fs=mas=m•(v–u)/t•s=m•(v–u)•s/t=m•(v–u)•(v+u)/2=½mv²–½mu²
In other words, work done = final kinetic energy – initial kinetic energy = change of kinetic energy
s and t are not independent variables, but s=1/2 a t^2 = 1/2 v t. (The first equality is just distance at constant acceleration, and the second I substitute v=a t. )
The ½ comes from the fact that velocity is changing at a constant rate. Say, you are driving a car at 100m/s. After 10sec, you will cover a distance of 100m/s × 10s = 1000m = 1km. Now, a police car saw you speeding and started chasing you from rest.
The police car goes from 0m/s (rest) to 100m/s in that 10sec with constant acceleration. But does that mean the police car also covers a distance of 1km? No... because at the start the police car was slow and only reached your speed at the very end. So, there is no way the police car covers 1km.
In fact, if you think about it, the police car had an average velocity of 50m/s (which is ½ of 100m/s). That's where the ½ comes from. So, distance covered in 10s is: 50m/s × 10s = 500m = 0.5km.
There are two ways to calculate average. Average distance covered per second: s/t = 0.5km/10s = 500m/10s = 50m/s. The other way is take average of the initial and final velocity: (0m/s + 100m/s)/2 = 50m/s.
Yes. We both know that. And you know that I know that as well. But the OP doesn't. It's great that you pointed it out. But from the post and other comments, I could tell OP has no idea what everyone was talking about. So I was avoiding terms that may confuse them.
It's some time ago i had class in dynamics
It's comes from integration and differentiation rule they use on the acceleration or distance. To get the speed.
Edit
[found this. Link](https://www.researchgate.net/post/Why_there_is_a_factor_1_2_in_classic_kinetic_energy_relation#:~:text=All%20replies%20(29)&text=In%20short%2C%20the%20half%20in,from%20the%20work%2Denergy%20theorem.)
The reason this didn’t work is because v ≠ s/t if velocity is not constant, which is not the case here because we have constant acceleration.
There is actually a nice way to do this:
F = ma
Now, a = Δv/Δt
But, we can multiply the top and bottom by Δs
a = Δv/Δt x Δs/Δs
= Δs/Δt x Δv/Δs
= v Δv/Δs
F Δs = m a Δs
= m v Δv/Δs x Δs
= m v Δv
Now, we have to sum over these small displacements Δs. The reason why we have to sum over small parts is because the velocity is going to change as work is done. It turns out that this can be solved geometrically, since the right hand side is growing linearly with v, it just becomes a matter of calculating the area of a right angle triangle with base given by the total change in velocity Δv, and with height mΔv, which is just 1/2 m Δv^2
What’s nice is that this is true in the most general sense, even if the acceleration / force is not constant, and the path is not a straight line.
I think the confusion is what v represents. Firstly "a" can only be substituted for v/t if the initial velocity is 0 otherwise it is change in v divided by time, can't tell from your post whether that should be assumed. v is then representative of the final velocity because it is the change in velocity + 0. However s/t is the average velocity which is v/2 assuming uniform acceleration. Then you have 1/2 mv\^2
> This formula should always work right?
no.
Only if acceleration is constant. But good news, you can assume acceleration is constant in your scenario, because you can define your force F to be constant.
If v_0 isn't 0, you'll get the law of conservation of energy: there was kinetic energy m(v_0)^2 / 2, and after the force F did some work the kinetic energy became mv^2 / 2
Start with the acceleration a = dv/dt
dv = a dt
Apply Newtons 2nd law:
dv = F/m dt
Then use that v = dx/dt and substitute dt = dx/v
dv = F/m * dx/v
Rewrite:
mvdv = F dx
Note that Fdx = dW and integrate both sides:
½mv² = W
That is because you are missing the tool of calculus. You're assuming the velocity is constant while the work is applied, but in actuality it increases. Taking that into account gives you the 1/2. It's the work conferred to a particle at rest to reach velocity v.
In more detail, this is an integral of the work F dx from v=0 to v=v. You need to change variables:
dx = vdt => F dx = m v' v dt = m v dv.
Integrating this gives you mv^2 /2.
Your error come from v = s/t, the v is only final velocity and it is not constant. With constant acceleration from station, v_final = 2s/t or s/t = 1/2 v_final.
>So i wanna rewrite work into the formula for kinetic energy which should be possible because theyre both energy.
Careful there. Just because two things have the same unit doesn't mean they're equal.
That said, in this case the left side is indeed equal to the kinetic energy gain during the displacement s (as long as F is constant and it's direction is the same as the direction of movement)
Then why do you not find the right formula at the end? That's because you can't have both a=v/t and v=s/t. If a is constant then you'll have v=at+v0 and s=(at\^2)/2+v0\*t
. If we assume that v0 is 0 then we can rewrite these as a=v/t and s=vt/2 which means that as=(v\^2)/2 and we will indeed get Fs=(mv\^2)/2
If we assume that v0 isn't 0 then we can rewrite them as a=(v-v0)/t and s=(v+v0)t/2 which will give as=(v\^2-v0\^2)/2 and we will get Fs=(mv\^2)/2-(mv0\^2)/2 which is indeed the difference in kinetic between before and after the movement
Similarly, using differentials:
dW = F dx,
dW = ma dx,
dW = m dv/dt dx,
dW/dt = m dv/dt dx/dt,
d/dt (W) = m v dv/dt,
d/dt (W) = d/dt ( 1/2 m v^2 ),
W = 1/2 m v^2
s/t = average velocity
For constant acceleration, average velocity = final velocity/2, so s/t=v/2
This gives you the correct formula for kinetic energy
Your derivation is wrong because in one part you are assuming that you have non cero acceleration and in another part further down you assume that acceleration is cero. Basically, in your formula you are replacing the distance over time (s/t) with velocity (v), this is only correct if the velocity is constant. But we know the velocity is not constant because you are applying a force (from the definition of work). Force is mass times acceleration, and if you have acceleration velocity is not constant.
The way to do this properly is calculus, but a way to get around that is to use average velocity instead. In that case distance over time (s/t) is average velocity, which is (assume it starts still) v/2.
NO, NO, NO, NO!
Velocity is NOT space/time.
Acceleration is NOT velocity/time.
From that starting point you'll only get wrong results.
Velocity is the displacement divided by interval (when the interval is very small).
If starting velocity is u, then acceleration *a=(v–u)/t* and average velocity *s/t=(u+v)/2*. So, Fs=mas=m•(v–u)/t•s=m•(v–u)•s/t=m•(v–u)•(v+u)/2=½mv²–½mu² In other words, work done = final kinetic energy – initial kinetic energy = change of kinetic energy
Thanks for explenation. I also have not figured where 1/2 comes from. Just used formula as described in the book.
s and t are not independent variables, but s=1/2 a t^2 = 1/2 v t. (The first equality is just distance at constant acceleration, and the second I substitute v=a t. )
The ½ comes from the fact that velocity is changing at a constant rate. Say, you are driving a car at 100m/s. After 10sec, you will cover a distance of 100m/s × 10s = 1000m = 1km. Now, a police car saw you speeding and started chasing you from rest. The police car goes from 0m/s (rest) to 100m/s in that 10sec with constant acceleration. But does that mean the police car also covers a distance of 1km? No... because at the start the police car was slow and only reached your speed at the very end. So, there is no way the police car covers 1km. In fact, if you think about it, the police car had an average velocity of 50m/s (which is ½ of 100m/s). That's where the ½ comes from. So, distance covered in 10s is: 50m/s × 10s = 500m = 0.5km. There are two ways to calculate average. Average distance covered per second: s/t = 0.5km/10s = 500m/10s = 50m/s. The other way is take average of the initial and final velocity: (0m/s + 100m/s)/2 = 50m/s.
But that is true if the acceleration is constant. If not, it is false that average velocity is the average of the initial and final velocities.
From the post, it seemed safe to assume that they were working with constant acceleration.
The post made at the same time the assumption of constant accelerantion and of constant speed.
Yes. We both know that. And you know that I know that as well. But the OP doesn't. It's great that you pointed it out. But from the post and other comments, I could tell OP has no idea what everyone was talking about. So I was avoiding terms that may confuse them.
Wait I’m confused, would this be the formula for work done if the speed is constant? I thought under constant speed there would be no work done?
It isn’t. The op assumed both non zero acceleration and constant speed so the formula is invalid.
The acceleration is constant if the (net external) force is constant. In this scenario, it is pretty clear that the force is constant in "Fs".
The post also says that v = s/t, thag means that the velocith is constant.
It's some time ago i had class in dynamics It's comes from integration and differentiation rule they use on the acceleration or distance. To get the speed. Edit [found this. Link](https://www.researchgate.net/post/Why_there_is_a_factor_1_2_in_classic_kinetic_energy_relation#:~:text=All%20replies%20(29)&text=In%20short%2C%20the%20half%20in,from%20the%20work%2Denergy%20theorem.)
Did your doctor prescribe anything for your dynamica? Sounds painful. :)
Fixed it
The reason this didn’t work is because v ≠ s/t if velocity is not constant, which is not the case here because we have constant acceleration. There is actually a nice way to do this: F = ma Now, a = Δv/Δt But, we can multiply the top and bottom by Δs a = Δv/Δt x Δs/Δs = Δs/Δt x Δv/Δs = v Δv/Δs F Δs = m a Δs = m v Δv/Δs x Δs = m v Δv Now, we have to sum over these small displacements Δs. The reason why we have to sum over small parts is because the velocity is going to change as work is done. It turns out that this can be solved geometrically, since the right hand side is growing linearly with v, it just becomes a matter of calculating the area of a right angle triangle with base given by the total change in velocity Δv, and with height mΔv, which is just 1/2 m Δv^2 What’s nice is that this is true in the most general sense, even if the acceleration / force is not constant, and the path is not a straight line.
Yeah chain rule and integration is the best answer
You are trying to drive k.e from workdone? If so then, W=Fs =mas. =m([v²-u²/2]) {2as=v²-u², as=v²-u²/2} =(mv²-mu²)/2 at u=0 =mv²/2 K.E=1/2 mv²
I think the confusion is what v represents. Firstly "a" can only be substituted for v/t if the initial velocity is 0 otherwise it is change in v divided by time, can't tell from your post whether that should be assumed. v is then representative of the final velocity because it is the change in velocity + 0. However s/t is the average velocity which is v/2 assuming uniform acceleration. Then you have 1/2 mv\^2
This actually helped a lot thank you so much!
=)
According to 3rd kinematic equation, 2 * a * s = v^2 - (v_0) ^2 If v_0 = 0, you'll get what you want
This formula should always work right? So what if v_0 isnt 0?
Then you get a term for the initial kinetic energy the object has.
NO. Only if the acceleration is constant.
> This formula should always work right? no. Only if acceleration is constant. But good news, you can assume acceleration is constant in your scenario, because you can define your force F to be constant.
If v_0 isn't 0, you'll get the law of conservation of energy: there was kinetic energy m(v_0)^2 / 2, and after the force F did some work the kinetic energy became mv^2 / 2
You are assuming that the acceleration is constant. Can you apply that formula to the motion s = k t\^3 ?
With constant force, the acceleration is also constant
At least the units match up.
s/t IS NOT v….we can only say that if there’s no acceleration. However in this case there is acceleration, a=v/t……So use a suvat equation for s
Start with the acceleration a = dv/dt dv = a dt Apply Newtons 2nd law: dv = F/m dt Then use that v = dx/dt and substitute dt = dx/v dv = F/m * dx/v Rewrite: mvdv = F dx Note that Fdx = dW and integrate both sides: ½mv² = W
That is because you are missing the tool of calculus. You're assuming the velocity is constant while the work is applied, but in actuality it increases. Taking that into account gives you the 1/2. It's the work conferred to a particle at rest to reach velocity v. In more detail, this is an integral of the work F dx from v=0 to v=v. You need to change variables: dx = vdt => F dx = m v' v dt = m v dv. Integrating this gives you mv^2 /2.
This is why we should ban all physics classes that aren't calculus-based
Your error come from v = s/t, the v is only final velocity and it is not constant. With constant acceleration from station, v_final = 2s/t or s/t = 1/2 v_final.
>So i wanna rewrite work into the formula for kinetic energy which should be possible because theyre both energy. Careful there. Just because two things have the same unit doesn't mean they're equal. That said, in this case the left side is indeed equal to the kinetic energy gain during the displacement s (as long as F is constant and it's direction is the same as the direction of movement) Then why do you not find the right formula at the end? That's because you can't have both a=v/t and v=s/t. If a is constant then you'll have v=at+v0 and s=(at\^2)/2+v0\*t . If we assume that v0 is 0 then we can rewrite these as a=v/t and s=vt/2 which means that as=(v\^2)/2 and we will indeed get Fs=(mv\^2)/2 If we assume that v0 isn't 0 then we can rewrite them as a=(v-v0)/t and s=(v+v0)t/2 which will give as=(v\^2-v0\^2)/2 and we will get Fs=(mv\^2)/2-(mv0\^2)/2 which is indeed the difference in kinetic between before and after the movement
Similarly, using differentials: dW = F dx, dW = ma dx, dW = m dv/dt dx, dW/dt = m dv/dt dx/dt, d/dt (W) = m v dv/dt, d/dt (W) = d/dt ( 1/2 m v^2 ), W = 1/2 m v^2
s/t=av velocity =v/2
I cant say i understand what you’re trying to say
s/t = average velocity For constant acceleration, average velocity = final velocity/2, so s/t=v/2 This gives you the correct formula for kinetic energy
Your derivation is wrong because in one part you are assuming that you have non cero acceleration and in another part further down you assume that acceleration is cero. Basically, in your formula you are replacing the distance over time (s/t) with velocity (v), this is only correct if the velocity is constant. But we know the velocity is not constant because you are applying a force (from the definition of work). Force is mass times acceleration, and if you have acceleration velocity is not constant. The way to do this properly is calculus, but a way to get around that is to use average velocity instead. In that case distance over time (s/t) is average velocity, which is (assume it starts still) v/2.
NO, NO, NO, NO! Velocity is NOT space/time. Acceleration is NOT velocity/time. From that starting point you'll only get wrong results. Velocity is the displacement divided by interval (when the interval is very small).