A number of identical cubical blocks of edge | vedclass

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(B) The total number of blocks is $N = n + 1$. Initially, all blocks are on the table, so their center of mass is at height $h_i = a/2$. After stackin

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$\frac{1}{2}ma\,g\left( {{n^2} + n} \right)$

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(B) The total number of blocks is $N = n + 1$. Initially, all blocks are on the table, so their center of mass is at height $h_i = a/2$. After stacking them into a column of height $H = (n + 1)a$, the center of mass of the column is at $h_f = H/2 = (n + 1)a/2$. The total mass of the system is $M = (n + 1)m$. The work done is equal to the change in gravitational potential energy: $W = \Delta U = M g (h_f - h_i)$. $W = (n + 1)m g \left[ \frac{(n + 1)a}{2} - \frac{a}{2} \right]$. $W = (n + 1)m g \left[ \frac{na + a - a}{2} \right] = (n + 1)m g \left( \frac{na}{2} \right)$. $W = \frac{1}{2} m a g (n^2 + n)$.

$A$ number of identical cubical blocks of edge $a$ and mass $m$ are lying on a horizontal table. The work done on the blocks in arranging them in a column of height $(n + 1)a$ on the table is

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