$A$ particle with constant total energy $E$ moves in one dimension in a region where the potential energy is $U(x)$. The speed of the particle is zero where

$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

$A$ chain of mass $M$ and length $L$ is placed on a table such that $L/4$ of its length hangs over the edge. What is the work done by an external force to pull the hanging part back onto the table?

In the equation $W = -mgh$,what does the negative sign indicate?

$A$ uniform chain of length $2\,m$ is kept on a table such that a length of $60\,cm$ hangs freely from the edge of the table. The total mass of the chain is $4\,kg$. What is the work done in pulling the entire chain onto the table? (Take $g = 10\,m/s^2$) ................ $J$

The potential energy of a particle of mass $5 \ kg$ moving in the $XY$ plane is given by $V = -7x + 24y$ joules,where $x$ and $y$ are in metres. Initially at $t = 0$,the particle is at the origin $(0, 0)$ moving with a velocity of $\vec{v}_0 = 6[0.24 \hat{i} + 0.7 \hat{j}] \ m/s = [1.44 \hat{i} + 4.2 \hat{j}] \ m/s$. Then: