Question

class 11th work, energy and power ​

Answer 1

$\Huge {\orange {\bf {\underline {\underline {ÀnSwer}}}:-}}$

The initial total potential energy of all the blocks lying on the ground is

$\sf U_{1} = n \times mg \frac{a}{2}$

The potential energy, when the block are piled to from a single column is

$\sf U_{1} = nmg \frac{na}{2} = {n}^{2} mg \frac{a}{2}$

$\therefore$ work done in arranging the blocks one above the other is

$\sf W = U_{2} - U_{1}$

$\sf W = {n}^{2} mg \frac{a}{2} - nmg \frac{a}{2}$

$\sf W = mg \frac{a}{2} ( {n}^{2} - n)$

$\boxed{ \sf W = mg \frac{a}{2} n(n - 1) }$