Question

Twenty identical cubical blocks each of mass 100gm and of side 10cm are lying on a horizontal surface. Work done in piling them one above the other is, $\left(g=10m{s}^{-2}\right)$

• A. 20 J
• B. 2 J
• C. 19 J
• D. 10 J

Answer 1

The correct option is C 19 J

Initially,

$U_p=\left(20\right)(-mgncom)5cm\updownarrow$

$=\left(20\right)(\frac{1}{10}\times 10\times \frac{5}{100})=1J$

After piling $U_p$,

$U_f=-m_{system}g{h}_{comofsystem}$

$U_f=(-20\times \frac{1}{10})\left(10\right)(5\times 20\times \frac{1}{100})$

$U_f=-20J$

$W=-\Delta UE$

$={-U}_f+U_p=-20+1=1$

Option$–c$ is correct.