# Work done and change in mechanical energy

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

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

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

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

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

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

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.