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

Consider ..n.. number of indentical cubes each of mass ..m.. and side

1.	Consider ..n.. number of indentical cubes each of mass ..m.. and side ..a.. lying on a level ground. Find work done in piling them one above the other.
Answer» <html><body><p></p>Solution :The initial total <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> energy of all the blocks lying on the ground is ` U_(1) = nxxmg""(a)/(2)` <br/> The potentia energy, when the blocks are piled to form a single <a href="https://interviewquestions.tuteehub.com/tag/column-922494" style="font-weight:bold;" target="_blank" title="Click to know more about COLUMN">COLUMN</a> is `U_(2) = nmg""(na)/(2) = <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> ^(2) <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>""(a)/(2)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Work done in arranging the blocks one above the other is `W = U_(2) - U_(1)` <br/> `W=n^(2)mg""(a)/(2)-nxxmg""(a)/(2)(n^(2)-n)` <br/> or `W = mg""(a)/(2)n (n-1)`</body></html>