# Work done by reaction forces between objects

Assume that there are no friction forces. If we had a particle sliding down a wedge that is free to move on a smooth surface, why do we ignore the work done by the reaction forces on both the particle and the wedge?

I would have thought $KE_\text{wedge} + KE_\text{particle} = W_{g\text{ on particle}} + W_{x\text{-component of reaction force on wedge}} + W_{x\text{-component of reaction force on particle}} + W_{y\text{-component of reaction force on particle}}$.

Instead, we find $KE_\text{wedge} + KE_\text{particle} = W_{g\text{ on particle}}$.

Why is the work done by the internal reaction forces zero?

Let's do the calculation in Vectors (you can also break them down into x and y direction, and get the same result)

1. Considering the particle, it feels two forces: the gravity $\vec G$ and the holding force by the wedge $\vec N$ (the so-called inner force); its velocity can be written as $\vec v(t)$.
2. Considering the wedge, it feels three forces: the gravity $\vec G'$, the compressing force by the particle $-\vec N$ (remember the third Newtonian law) and the holding force by the ground $\vec N'$ (the so-called inner force); its velocity can be written as $\vec v'(t)$.
3. Considering the wedge's shape does not change, the relative velocity between the particle and the wedge should be along the wedge's face: $\left[\vec v(t)-\vec v'(t)\right]\cdot \vec n = 0$, where $\vec n$ is the normal vector of the wedge face (say, $\vec n = -\sin\theta\ \vec e_x + \cos \theta\ \vec e_y$)
4. The compressing & holding forces between the particle and the wedge should be perpendicular to the wedge face, that is: $\vec N =k \cdot \vec n,\ k\in R$.
5. Now comes to the final calculation: For the total work done by the internal forces on two subjects(particle+wedge) $W$, we have $P = \dfrac{{\rm d} W}{{\rm d} t} = \vec N\cdot \vec v(t) + (-\vec N)\cdot \vec v'(t) = \vec N \cdot(\vec v(t)-\vec v'(t)) = k\cdot \vec n \cdot \left[\vec v(t)-\vec v'(t)\right] = k \cdot 0$ so $W = \int_0^t \left(\dfrac{{\rm d} W}{{\rm d} t}\right) {\rm d} t = 0$

Well whenever using the work energy theorem always remember that

Wc+Wnc+Wext=$\Delta$KE

where Wc is the work done by conservative internal forces

Wnc is the work done by non-conservative internal forces

Wext is the work done by external forces

Now here non-conservative forces (such as friction,viscous force etc.) are not present,so work done by them will be 0.

Here the other 2 forces present are weight and normal reaction force. According to the definition of work, $W=\vec{F}\cdot\vec{s}$, i.e the dot product of force and displacement. Since here work displacement of the block is always parallel to the incline and Normal reaction force is perpendicular to the incline.Therefore the work done by it will be 0. Hence only the weight will be doing work and that too its component which is parallel to the displacement.