Question

$n$ identical cubes each of mass $m$ and edge $L$ are lying on a floor. If the cubes are to be arranged as one over the other in a vertical stack, the work to be done is:

• A. $\frac{Lmng(n-1)}{2}$
• B. $\frac{Lg(n-1)}{mn}$
• C. $\frac{(n-1)}{Lmng}$
• D. $\frac{Lmng}{2(n-1)}$

Answer 1

The correct option is A $\frac{Lmng(n-1)}{2}$

Work done $=\Delta (P.E)$
Initial $P.E.$ $=n\times m\times g\times \frac{L}{2}$
Final $P.E.$ $=n\times m\times g\times \frac{nL}{2}$
Work done $=\Delta (P.E.)$ $=nmg\frac{L}{2}(n-1)$