Question

A rectangular block of dimensions
10m x 6m x 3m is lying on a horizontal
surface with its smallest area in
contact with the surface. If the work
done in arranging it with its largest
area in contact with the surface is
13.86MJ, the density of the body is
(g=10ms-2)

​

Answer 1

The PE of the system when the block is lying on smallest surface is =mgh=m×10×5=50m

The PE of the system when the block is lying on largest surface is =mgh=m×10×1.5=15m

The work done in the process =50m−15m=35m

Now,

35m=13.86×106⇒m=396×103 kg

The density of the block

=mV=396×103100×60×30=2.2 gm/cc

Answer 2

Explanation:

see the above one picture