Question

An $8\text{kg}$ metal block of dimension $16\text{cm}\times 8\text{cm}\times 6\text{cm}$ is lying on a table with its face of largest area touching the table. If $g=10{\text{ms}}^{-2},$ the minimum amount of work done (in Joules, upto two decimals ) in making it stand with maximum length vertical is

Answer 1

Given
$M=8\text{kg}$
Dimension of mass $=16\times 8\times 6\text{cm}$
Area of largest face is $=16\times 8$.
Initial height of center of mass
$h_1=\frac{6\text{cm}}{2}=3\text{cm}=0.03m$
Final height of center of mass
$h_2=\frac{16}{2}\text{cm}=8\text{cm}=0.08m$
Work done from shifting centre of mass from $0.03\text{m}$ to $0.08\text{m}$
$W=\text{mg}\Delta h$
$W=mg(h_2-h_1)$
$=8\times 10(0.08-0.03)$
$=8\times 10\left(0.05\right)=4\text{J}$