Consider ..n.. number of indentical cubes each of mass ..m.. and side ..a.. lying on a level ground. Find the work done in piling them one above the other.

Explanation

The initial total potential energy of all the blocks lying on the ground is ` U_(1) = nxxmg""(a)/(2)`
The potentia energy, when the blocks are piled to form a single column is `U_(2) = nmg""(na)/(2) = n ^(2) mg""(a)/(2)`
`therefore` Work done in arranging the blocks one above the other is `W = U_(2) - U_(1)`
`W=n^(2)mg""(a)/(2)-nxxmg""(a)/(2)(n^(2)-n)`
or `W = mg""(a)/(2)n (n-1)`