I have a doubt in this question, Diagram given below
In the Question it is asked that "calculate the final velocity of the block in the figure" and in the solutions it is given that the work done by normal force exerted by the surface is not considered when applying the work-energy theorem
but according to the proof of the work-energy theorem, the "'net' external force X displacement upto which the force was applied is equal to the change in kinetic energy of the object."
(could use integration by taking the variable force to be constant for infinitely small time intervals)
but by this definition normal force should also be added to calculate the net external force on the object as it also always acts in opposite directions to gravity (with different magnitudes obviously)
So why is the normal force not considered here? Is the question/solution wrong?
If it is correct, why? And what is the mathematical proof of not considering the normal force when applying the work-energy theorem?
Best Answer
You could indeed use the work-energy theorem to argue that the energy gained by the block is equal to the work done on it by the net external force acting on it, which is its weight $m\vec g$ plus the normal force $\vec N$. Note that both $\vec g$ and $\vec N$ are vectors and do not act in the same direction so you need to add them as vectors. Also note that the normal force acts perpendicularly to the surface, not in the opposite direction to gravity. So the net external force on the block is
$\vec F = m \vec g + \vec N$
The work done on the block when it moves a small distance $\vec {dx}$ is then
$dW = \vec F . \vec {dx} = (m \vec g + \vec N) . \vec {dx}$
Once again, note that $\vec {dx}$ is a vector, and to find the scalar quantity $dW$ then we must take the dot product or scalar product of $m \vec g + \vec N$ with $\vec {dx}$. Finally, we can integrate $dW$ across the whole motion to get
$\displaystyle W = \int dW = \int (m \vec g + \vec N) . \vec {dx}$
However, it is much simpler (and gives the same result) if we use conservation of energy to argue that the kinetic energy gained by the block is equal to the potential energy that it loses, which is $mgh$ where $h$ is the vertical distance that it has fallen.
To see that these two methods give the same result, notice that $\vec N$ is always perpendicular to $\vec {dx}$, so $\vec N . \vec {dx} = 0$ (in other words, the normal force does no work on the block) and we have
$\displaystyle W = \int (m \vec g + \vec N) . \vec {dx} = \int m \vec g . \vec {dx} + \int \vec N . \vec {dx} = m \int \vec g . \vec {dx} = mgh$