Question

A rectangular block 5m x 4m x 2m lies on a table with its largest surface in contact with the table. The work done to keep it so that the block rests on the smallest surface is, if its density is 600 k $m^{-3}$

• A. $352800J$
• B. zero
• C. $376000J$
• D. $24,0000J$

Answer 1

The correct option is A $352800J$

The volume of Rectangular block $=5m\times 4m\times 2m$

$=40m^3$

Mass = Volume $\times$ Density

$\begin{array}{c}=40m^3\times 600k{m}^{-3}\\ =24,000=2.4\times {10}^{4}hg\end{array}$

Potential energy of block when it is lying on its largest surface $=mgh$

$=2.4\times {10}^4\times 9.8\times 1joule$

Potential energy of the block when it is lying on

$=mgh=2.4\times {10}^4\times 9.8\times 2.5$

Work done = difference in potential energy

$\begin{array}{c}=\left(2.4\times {10}^4\times 9.8\times 2.5-2.4\times {10}^4\times 9.8\times 1\right)joule\\ =2.4\times {10}^4\times 9.8\times 1.5joule\\ =352800joule\end{array}$

Option A