An 8 kg metal block of dimensions 16 cm × 8 cm × 6 cm is lying on a table with its face of largest area touching the table. If g = 10 ms - 2 , then the minimum amount of work done in making it stand with its length vertical is

Step-by-step solution

When face of the largest area is touching the table, height of centre of gravity above the table h ⁡ 1 = 6 2 = 3 cm With its length vertical, height of centre of gravity would become h ⁡ 2 = 1 6 2 = 8 cm Minimum work required W = (P.E.) 2 – (P.E.) 1 = mg(h 2 – h 1 ) W ⁡ = 8 × 1 0 8 - 3 1 0 0 = 4 J