Question

$n$ identical cubes each of mass $m$ and side $l$ are on a horizontal surface. Then the minimum amount of work done to arrange them one over another in a vertical stack is

• A. $nmgl$
• B. $\frac{mgl{n}^2}{2}$
• C. $\frac{mgln(n-1)}{2}$
• D. $\frac{mgln(n+1)}{2}$

Answer 1

The correct option is C $\frac{mgln(n-1)}{2}$



Initially, the blocks are lying on the horizontal surface. The center of mass of each block will be at a length $\frac{l}{2}$ from the horizontal. Thus, the potential energy of each block will be $\frac{mgl}{2}$.

Thus, the initial potential energy of the blocks is

$P{E}_1=nmg\frac{l}{2}$

Now, when the blocks are stacked together the total mass becomes $nm$ and the total length is $nl$, so the center of mass lies at $n\frac{l}{2}$

Thus, the final PE of the system is

$P{E}_2=nmg\left(\frac{nl}{2}\right)$

Now, we can calculate the work done.

Work done $W=W_2-W_1$

$W=P{E}_2-P{E}_1$

$W=nmg\left(\frac{nl}{2}\right)-nmg\left(\frac{l}{2}\right)$

$W=\frac{nmgl}{2}[n-1]$

$W=\frac{mgl}{2}n(n-1)$