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 m s -2, the minimum amount of work done in making it stand with its length vertical is

Question

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 m s -2, the minimum amount of work done in making it stand with its length vertical is (A) 0.4 J (B) 6.4 J (C) 4 J (D) 12.8 J

Tardigrade Admin · Accepted Answer

When face of the largest area is touching the table, height of centre of gravity above the table

h_{1} = \frac{6}{2} = 3 c m

With its length vertical, height of centre of gravity would become

h_{2} = \frac{16}{2} = 8 c m

Minimum work required

W = (P . E .)_{2} - (P . E .)_{1} = m g (h_{2} - h_{1})

∴ W = \frac{8 \times 10 ( 8 - 3 )}{100} = 4 J

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

0.4 J

6.4 J

4 J

12.8 J

When face of the largest area is touching the table, height of centre of gravity above the table h1​=26​=3cm With its length vertical, height of centre of gravity would become h2​=216​=8cm Minimum work required W=(P.E.)2​−(P.E.)1​=mg(h2​−h1​) ∴W=1008×10(8−3)​=4J

Q. An $8 k g$ metal block of dimensions $16 c m \times 8 c m \times 6 c m$ is lying on a table with its face of largest area touching the table. If $g = 10 m s^{- 2}$ , the minimum amount of work done in making it stand with its length vertical is