First slide

Work done by Diff type of forces

Question

‘n’ identical cubes each of mass ‘m’ and side ‘l’ are on the horizontal surface. Then the minimum amount of work done to arrange one on the other is

Moderate

A

nmgl
B

$\frac{m g l n^{2}}{2}$
C

$\frac{m g \ln (n -!)}{2}$
D

$\frac{m g \ln (n +!)}{2}$

Solution

work done w= mgh
$Work done to put the second one on the top of the first mg \{l + \frac{l}{2}\} - mg \frac{l}{2} = mgl (center of mass of the cube has to be considered as the position of the center of mass) similarly for the third one to put on the second work done is mg \{2 l + \frac{l}{2}\} - mg \frac{l}{2 = 2 mgl} similarly for the next work done is 3 mgl and so on upto n - 1 terms so total work done is mgl (1 + 2 + 3 + . . . . . . . (n - 1)) \frac{mgln (n - 1)}{2}$
=

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