I really need some help with a physics problem, and I guess my doubt is more conceptual, than question based. But still, let me pose it to you:
A uniform chain of mass $M$ and length $L$ lies on a rough table, with coefficient of friction $k$. When 1/3rd part of the chain is hanging off the edge, it starts slipping. Find the work done by the frictional force on the chain, at the instant the last link falls off.
Now, to find the work, I know we must use integration. The work done would be
$$\int k \, p \, T \, g\, dp \tag{1}$$
Where $k$ is the coefficient of friction, $p$ is the varying length, $T$ is the mass per unit length ($= \frac{M}{l}$), and $g$ is the acceleration due to gravity. After applying appropriate limits, we get the answer.
The integrand in $(1)$ simply should mean force $\times$ displacement.
Now does my question come in.
In the integrand, the Force quantity is defined by the variable frictional quantity, which is equal to $(k \, p \, T \, g)$, so where did the quantity of displacement of the chain go? Shouldn't there be an extra $p$ in the intgrand, such that it becomes $(k \, p^2 \, T \, g)$?
Or is the notation $dp$ itself used to signify the displacement?
I know the question is a bit complicated, but please help me out. It will be highly appreciated.
Best Answer
The quantity of displacement that you are looking for is $dp$ in the integral.
At any instant the friction on the chain is $kpTg$, instantaneous work done by friction when the chain moves $dp$ length would be $-kpTg \times dp$ (force x displacement). Negative sign because work is done against direction of motion or because friction is opposing motion.
To find out the total work done against friction, integrate $-kpTgdp$ with $p$ varying from $2l/3$ to zero, here $p$ is the length of the chain on the table.
On integrating, the expression that you get should be $$-\frac{k(p^2)Tg}2$$ Put the limits from $2l/3$ to zero.