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


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


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

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


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

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:


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

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

$W=\vec{F}\cdot \vec s$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.