# General relationship between work and energy [closed]

I am messing around with formulae and I come across things that don't seem consistent. I would like someone to point out where I make a mistake.

Starting with the formula for gravitational potential energy where we define the ground as $0$. $$Gpe=mgh$$ Which can be derived from $$W=Fs$$ since $$F=ma$$ therefore subbing in $$W=(ma)s$$ and letting $a=g$ and $s=h$ then we get to our solution that $Work(gpe)=mgh$

Since this is mathematically accurate then lets say I want to use the better definition of force.... $$F={dP\over dt}$$ Which can be rewritten as $P=mv$ therefore $$F={{{m\Delta v}\over\Delta t}}$$ Which means that we can equate $W=Fs$ with $F={{{m\Delta v}\over\Delta t}}$ which gives us the equation of: $$W={{m\Delta v}\over\Delta t}{s\over 1}$$ in which we can see that ${s\over \Delta t}=v$ and therefore subbing in we achieve: $$W=m \Delta v v$$ Which is really really close to $$W={1\over2}m v^2$$ EXCEPT the $1\over 2$ is missing and I am not sure about if $\Delta v v$ can be written as $v^2$

I am sure people are going to show me the integration and how the half constant comes in by taking the integral of velocity with respect to velocity after changing the $$W=\int Fdx$$ to $$W=m\int vdv$$ But I understand this is a way to get the half constant... But MY REQUEST is for someone to show me how to achieve that result by equating expressions in the algebra style method I did. I badly need someone to point out any mistakes and show me how to rectify them.. obviously there are a few

## 2 Answers

The mistake is in: ${s\over \Delta t}=v$, because the speed is increasing over time and $v$ is the final speed.

Assume an object is standing still and a constant force $F$ is applied over distance $s$, ending with speed $v$.

If you start with $F=ma$ then: $Fs=mas$.

Given that the acceleration is constant, we can express $s$ in the ultimate speed $v$ as $s=\frac{1}{2}vt$, the average speed times t and we also have $a=\frac{v}{t}$. So that gives:

$Fs=mas=m\frac{v}{t}\frac{1}{2}vt=\frac{1}{2}mv^2$

Using the terminoligy in the question if we start from:

$W={{m\Delta v}\over\Delta t}{s\over 1}$

Then $s = \frac{1}{2}\Delta v \Delta t$, because $\frac{1}{2}\Delta v$ is the average speed over the time period.

If we fill this in:

$W={{m\Delta v}\over\Delta t}\frac{1}{2}\Delta v \Delta t= \frac{1}{2}m(\Delta v)^2$

• It gets really confusing if you don't start with F=ma but instead use F=rate of change of momentum. I think your answer is great but I want someone to do this without using F=ma, because I want to wrap my head around the different "v"s and what they mean without simplifying any into "a"s Commented Feb 16, 2018 at 9:50
• I know it seems redundant but it will help me and others with conceptual understanding Commented Feb 16, 2018 at 9:51
• Ok, I added that to the answer. Commented Feb 16, 2018 at 10:06
• Ahhhh yes.. So delta V is actually just V(final)-U(inital speed) and that both "V"s are actually "delta Vs" and so the equation can be written as (delta v)^2 And I guess you assumed the object begun at rest. I think I get it although there is one thing that I feel like i don't appreciate enough... Commented Feb 16, 2018 at 10:14
• And that would be when to use "V" and when to use "delta V" because I understand there are times where both are used in the same equation I am pretty sure Commented Feb 16, 2018 at 10:15

the v in 1/2 mv^2 your final expression is final velocity And displacement over time does not give final velocity It rather gives avarage velocity which if the stiuation is asuumed to be uniformly accelerated is V(initial) +V(final)÷2 (Average of two intial and final velocity) And your delta v will become v final_v(initial)(difference of initial and final velocity) This gives the product v*delta v as difference of squares of the final and intial velocity divided by two(v(final)^2_v(initial)^2)/2 Rest is a peice of cake multiply with m Get Final kinetic energy MINUS initial kinetic energy If you asume particle starts from rest Then you get W=1/2MV^2 if you find any difficulty still I will paste an image latter on

Allaha bless you