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I have a hard time understanding the equation $W_{Driving} +W_{Friction} = \Delta E_{mechanical}$ Is there any derivation of it?

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First considering a box on a horizontal ground initially moving with speed $u$ on a rough surface and then a force is applied to the box in the direction of movement that causes the box to accelerate and reach a speed $v$ after travelling a distance $s$. So

$v^2=u^2+2as$

$a = \dfrac{F_{Net}}{m}$

$F_{Net}= F_{D}-F_{F}$

$v^2=u^2+2\dfrac{F_D.s-F_F.s}{m}$

$W_D = F_D.s, W_F = \vec F_F\cdot\vec s = -F_F.s$

finally, we get to desired equation $W_D+W_F=\dfrac{mv^2}{2}-\dfrac{mu^2}{2} = \Delta E_M$

Then try doing the exact thing but also adding in potential energy. Think of applying a driving force on a box which is moving with an initial speed of $u$ and reaching a speed $v$ by travelling $s$m against gravity on earth:

$v^2=u^2+2\dfrac{F_{Net}}{m}s$

$F_{Net}=F_D-F_{Gravity}-F_{Air Resistance}$

$v^2=u^2+2\dfrac{F_D.s-F_{Gravity}.s-F_{Air Resistance}.s}{m}$

$W=\vec{F}\cdot \vec s$ so $W_D = F_D.s, W_{Air Resistance} = -F_{Air Resistance}.s , W_G = -F_G.s$

$\dfrac{mv^2}{2}-\dfrac{mu^2}{2} = W_D + W_{Air Resistance}+W_G$

$\dfrac{mv^2}{2}-\dfrac{mu^2}{2} = \Delta E_K$

$W_G = -\Delta U $ because it is work done by a conservative force

$\Delta E_K = W_D + W_{Air Resistance}-\Delta U$

$W_D + W_{Air Resistance} = \Delta E_K + \Delta U$

$\Delta E_K + \Delta U = \Delta E_M$

So finally the equation: $W_D + W_{Air Resistance} = \Delta E_M$

I came up with an answer while writing the question so I post it for anyone that may have the same qs in the future.If anyone spots a mistake feel free to correct it.

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